C语言程序的设计习题参考答案(第二版_杜友福)

1 【C语言】《C语言程序设计教程(第二版)》习题答案说明1. 本文所指的《C语言程序设计教程(第二版)》是李凤霞主编、北京理工大学出版社出版的，绿皮。

2 第1章程序设计基础知识一、单项选择题(第23页)1-4.CBBC 5-8.DACA二、填空题(第24页)1.判断条件2.面向过程编程3.结构化4.程序5.面向对象的程序设计语言7.有穷性8.直到型循环9.算法10.可读性11.模块化12.对问题的分析和模块的划分三、应用题(第24页)2.源程序：main(){int i,j,k; /* i:公鸡数,j:母鸡数,k:小鸡数的1/3 */printf("cock hen chick\n");for(i=1;i<=20;i++)for(j=1;j<=33;j++)for(k=1;k<=33;k++)if (i+j+k*3==100&&i*5+j*3+k==100)printf(" %d %d %d\n",i,j,k*3);}执行结果：cock hen chick4 18 788 11 8112 4 843.现计算斐波那契数列的前20项。

递推法源程序：main(){long a,b;int i;a=b=1;for(i=1;i<=10;i++) /*要计算前30项,把10改为15。

*/{printf("%8ld%8ld",a,b);a=a+b;b=b+a;}}递归法源程序：main(){int i;for(i=0;i<=19;i++)printf("%8d",fib(i));}fib(int i){return(i<=1?1:fib(i-1)+fib(i-2));}执行结果：1 123 5 8 13 21 34 5589 144 233 377 610 987 1597 2584 4181 6765 4.源程序：#include "math.h";main(){double x,x0,deltax;x=1.5;do {x0=pow(x+1,1./3);deltax=fabs(x0-x);x=x0;}while(deltax>1e-12);printf("%.10f\n",x);}执行结果：1.32471795725.源程序略。

C语言程序‎设计习题参‎考答案习题 1一、判断题1．在计算机中‎，小数点和正‎负号都有专‎用部件来保‎存和表示。

2．二进制是由‎0和1两个‎数字组成的‎进制方式。

3．二进制数的‎逻辑运算是‎按位进行的‎，位与位之间‎没有进位和‎借位的关系‎。

4．在整数的二‎进制表示方‎法中，0的原码、反码都有两‎种形式。

5．有符号数有‎三种表示法‎：原码、反码和补码‎。

6．常用字符的‎A S CII‎码值从小到‎大的排列规‎律是：空格、阿拉伯数字‎、大写英文字‎母、小写英文字‎母。

解：1．F2．T 3．T 4．T 5．T 6．T二、单选题1．在计算机中‎，最适合进行‎数值加减运‎算的数值编‎码是。

A. 原码B. 反码C. 补码D. 移码2．已知英文小‎写字母m的‎A SCII‎码为十进制‎数109，则英文小写‎字母y的A‎S CII码‎为十进制数‎。

A. 112B. 120C. 121D. 1223．关于ASC‎II码，在计算机中‎的表示方法‎准确地描述‎是。

A. 使用8位二‎进制数，最右边一位‎为1B. 使用8位二‎进制数，最左边一位‎为1C. 使用8位二‎进制数，最右边一位‎为0D. 使用8位二‎进制数，最左边一位‎为04．设在机器字‎长4位，X＝0111B‎，Y＝1011B‎，则下列逻辑‎运算中，正确的是_‎_____‎_____‎。

A. X∧Y＝1000B. X∨Y＝1111C. X⊕Y＝0011D. ¯Y＝10005．下列叙述中‎正确的是（）。

A．高级语言就‎是机器语言‎B．汇编语言程‎序、高级语言程‎序都是计算‎机程序，但只有机器‎语言程序才‎是计算机可‎以直接识别‎并执行的程‎序C．C语言因为‎具有汇编语‎言的一些特‎性，所以是汇编‎语言的一种‎D．C源程序经‎过编译、连接，若正确，执行后就能‎得到正确的‎运行结果6．用C语言编‎写的源程序‎经过编译后‎，若没有产生‎编译错误，则系统将（）。

参考答案习题二(P33)一．单选题1.C2.B3.D4.C5.A6.D7.D8.B9.B 10.D 11.D 12.C 13.C . 15. A 14题最后一句应为printf("%f\n",d*y);结果为2.2二．填空题1. 182. int float double3. 10 114. 八十六十5. %三．阅读程序题1．10,10,9,102．j=1,i=2k=3,i=3j=3,i=2k=1,i=1习题三(P52)一．单选题1.D2.C3.D4.B5.A6.B7.C8.A9.C 10.B 11. -1,37777777777,4294967295 二．填空题5. L6. -1三．阅读程序题1. 6 6 6.00 6.002. x=127,x= 127,x=177,x=7fY=123.4567 , y= 123.46 , y=123.456703. 2,14. 12345. 4,36. -6,-6四．程序设计题1．#include "stdio.h"#include "math.h"main(){float a,b,c,d,x1,x2;a=2;b=-3;c=-5;d=b*b-4*a*c;x1=(-b+sqrt(d))/(2*a);x2=(-b-sqrt(d))/(2*a);printf("x1=%.2f,x2=%.2f\n",x1,x2);}2.#include <stdio.h>main(){ float a,v,s;scanf("%f",&a);v=a*a*a;s=6*a*a;printf("v=%.2f,s=%.2f\n",v,s);}3.#include <stdio.h>main(){ int a,b,c,t;scanf("%d%d%d",&a,&b,&c);printf("a=%d,b=%d,c=%d\n",a,b,c);t=c;c=b;b=a;a=t;printf("a=%d,b=%d,c=%d\n",a,b,c);}4.#include <stdio.h>main(){ char s1,s2;s1=getchar();s2=s1-32;printf("%c\n",s2);}习题四(P70)一．单选题1.C2.D3.D4.B5.A6.D7.D8.B9.C 10.A 11.B 12.D 13.passwarnerror .14.C 15. C 16.B 17. B 18. C二．填空题1. 非0 02. k==03. n%7==0 && n%8==0 else三．阅读程序题1. a=1,b=02. c=1四．程序设计题1.#include <stdio.h>main(){ int a,b;char c;printf("INPUT A+(-*/)B\n");scanf("%d%c%d",&a,&c,&b);switch(c){case'+':printf("%d+%d=%d\n",a,b,a+b);break;case'-':printf("%d-%d=%d\n",a,b,a-b);break;case'*':printf("%d*%d=%d\n",a,b,a*b);break;case'/':printf("%d/%d=%d\n",a,b,a/b);break;default:printf("INPUT ERROR!");}}2.#include <stdio.h>main(){ float x,y;scanf("%f",&x);if(x==0||x==2)y=0;else if(x>0) y=(x+1)/(x-2);else y=(x-1)/(x-2);printf("y=%f\n",y);}3.#include <stdio.h>main(){ int g;printf("请输入学生成绩:");scanf("%d",&g);printf("g=%d:",g);switch(g/10){case 10:printf("A\n");break;case 9:printf("B\n");break;case 8:printf("C\n");break;case 7:printf("D\n");break;case 6:printf("E\n");break;default:printf("F\n");}}4.#include <stdio.h>main(){ int x,n=0;scanf("%d",&x);if(x<=0||x>=10000)printf("Data Error!\n");else{if(x<10)n=1;else if(x<100)n=2;else if(x<1000)n=3;else n=4;}printf("n=%d\n",n);}习题五(P87)一．单选题1.B2.D3.C4.C5.B6.C7.D8.C9.A 10.D 11.B 12.B 13.B 14.D 15. C 二．填空题1. for语句while语句do-while语句2. a=14,y=263. k=14,n= -14. 395. s=196. 222227. 108. 79. 810. 5,5三．改错题应改为for（i=0;i<5;i++）j++;应改为int j=0;while(j<10){j++;i=j;} 注while(j<10)后不应有分号应改为while(j<10); 注while(j<10)后分号不能少s*=i;i++;应改为{s*=i;i++;}continue 应改为break四．程序设计题1.#include <stdio.h>main(){ int i,j,s=0;for(i=1,j=1;i<=100;i++,j=j*(-1))s=s+i*j;printf("s=%d\n",s);}1.（另一做法）#include<stdio.h>main(){int i,s;for(i=1,s=0;i<=100;i++)if(i%2==0)s=s-i;else s=s+i;printf ("%d\n",s);}2．#include <stdio.h>main(){ int i,j,s1=0,s2=0;for(i=0;i<10;i++){scanf("%d",&j);if(j>0)s1=s1+j;else s2=s2+j;}printf("s1=%d,s2=%d\n",s1,s2);}3．#include<stdio.h>main(){int i,s;for(i=6,s=0;i<=96;i=i+1)if(i%10==6||i/10==6)s=s+i;printf ("%d\n",s);}4．#include<stdio.h>main(){int i,a,b,c;for(i=100;i<=999;i++){a=(i%10);b=(i/10%10);c=(i/100);if(i==a*a*a+b*b*b+c*c*c)printf ("%d\n",i);}}5．#include <stdio.h>main( ){int i,j,k=0;for (i=1;i<=4;i++){for (j=1;j<=i;j++){k++;printf("%d",k%10);}printf("\n");}}6.#include <stdio.h>main( ){int i,j,k=0;for (i=-3;i<=3;i++){if(i<0)k=-i;else k=i;for(j=1;j<=k;j++)printf(" ");for(j=1;j<=7-2*k;j++)printf("*");printf("\n");}}习题六(P111)一．单选题1.D2.B3.C4.C5.C6.D7.A8.B二．填空题1. 20 0 192. 数组名3. 越界4. 65. j==k a[j][k]=1; a[j][k]=0;三．阅读程序题1． 6 5 43 2 12．aaabbbccc ddd3．2,2,1四．程序设计题1.#include<stdio.h>main(){ int a[4][4]={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16}; int i,j,s=0;for(i=0;i<4;i++)for(j=0;j<4;j++)if(i==j||i+j==3)s+=a[i][j];printf("%d",s);}2.#include <stdio.h>{ char a[80];int i,j=5; /*假设删除位置为5*/gets(a);for(i=j-1;a[i]!='\0';i++)a[i]=a[i+1];a[i]='\0';puts(a);}3．#include <stdio.h>#include <string.h>main(){ char a[80];int i,j=5; /*假设插入位置为5*/char s='t'; /*假设插入字符为t*/gets(a);for(i=strlen(a);i>j;i--)a[i+1]=a[i];a[j]='t';puts(a);}4．#include<stdio.h>main(){ int a[3][5]={1,3,5,7,9,2,4,6,8,10,3,5,8,7,6}; int i,j,s1[3]={0},s2[5]={0};for(i=0;i<3;i++)for(j=0;j<5;j++)s1[i]+=a[i][ j];for(i=0;i<5;i++)for(j=0;j<3;j++)s2[i]+=a[ j][i];for(i=0;i<3;i++){for(j=0;j<5;j++)printf("%6d",a[i][j]);printf (" |%6d\n",s1[i]);}for(i=0;i<33;i++) printf("-");printf("\n");for(i=0;i<5;i++) printf("%6d",s2[i]);printf("\n");}5．#include<stdio.h>{ char s[3][80];int a=0,b=0,c=0,d=0,e=0,i,j;for(i=0;i<3;i++) gets(s[i]);for(i=0;i<3;i++)for(j=0;s[i][j]!='\0';j++)if(s[i][j]>='A'&&s[i][j]<='Z')a++;else if(s[i][j]>='a'&&s[i][j]<='z')b++;else if(s[i][j]>='0'&&s[i][j]<='9')c++;else if(s[i][j]==32)d++;else e++;printf("%3d%3d%3d%3d%3d",a,b,c,d,e);}习题七(P145)一．单选题1.A2.B3.C4.A5.C6.D7.C8.B9.B 10.C 11.D 二．填空题1. 该函数内局部2. 整型3. k<=breturn y;4. x[i]return (ave);fun(a,20)5. 1;add(n-1);add(n);6. n*f(n-1)0;f(i)三．阅读程序，写出运行结果1．10,20,302．643．84. hlo5. sum=55四．程序设计题1.#include <stdio.h>main(){float add(float, float), sub(float, float);float aver(float, float),a,b;scanf("%f,%f",&a,&b);printf("add=%f, sub=%f\n ", add(a,b), sub(a,b)); printf("aver =%f", aver(a,b));}float add(float x, float y) {return(x+y);}float sub(float x, float y) {return(x-y);}float aver(float x, float y) {return((x+y)/2);} 2.#include<stdio.h>main(){ int a,b,c,abmax(int,int);scanf("%d%d%d",&a,&b,&c);printf("max=%d\n",abmax(abmax(a,b),c)); }int abmax(int a,int b){if(a>b)return a;else return b;}3.#include <stdio.h>main(){float x,fun(float, int);int n;scanf("%f%d",&x,&n);printf("%f\n", fun(x,n));}float fun(float x, int n){float y;if(n==0)y=1;else y=x*fun(x,n-1);return y;}4.#include <stdio.h>main(){long f(int);int n,k;scanf("%d",&n);for(k=1;k<=n;k++)printf("%ld,",f(k));printf("\n");}long f(int n){long y;if(n<=2)y=1;else y=f(n-1)+f(n-2); return y;}5.#include <stdio.h>#include <string.h> main(){char a[50];int n;void fun(char x[ ],int n); gets(a);n=strlen(a); fun(a,n);puts(a);}void fun(char a[ ],int n) {int k,s;for(k=0;k<n/2;k++) {s=a[k];a[k]=a[n-k-1];a[n-k-1]=s;}}5. （另一做法）#include <stdio.h>#include <string.h> main(){char a[50];int n;void fun(char x[ ],int n); gets(a);n=strlen(a); fun(a,n);puts(a);}void fun(char a[ ],int n) {int k;char b[50];for(k=0;k<n;k++)b[n-1-k]=a[k];b[k]=a[k];strcpy(a,b);}6.#include <stdio.h>#define KK 100main(){char a[KK];long sjz(char a[]);int i=0,f1=0;printf("input a data:");gets(a);for(i=0;a[i]!='\0';i++){if(a[i]>='0'&&a[i]<='9'||a[i]>='A'&&a[i]<='F'||a[i]>='a'&&a[i]<='f') continue;else {f1=1;break;}}a[i]='\0';if(f1==1)printf("Data Error!\n");else printf("result is :%d\n",sjz(a));}long sjz(char a[]){long n=0,i;;for(i=0;a[i]!='\0';i++){if(a[i]>='0'&&a[i]<='9')n=n*16+a[i]-'0';if(a[i]>='A'&&a[i]<='F')n=n*16+a[i]-'A'+10;if(a[i]>='a'&&a[i]<='f')n=n*16+a[i]-'a'+10;}return n;}6.（另一做法）#include <stdio.h>#define KK 100main(){char str[KK],c;long sjz(char a[]);int i=0,f1=0;printf("input a data:");while((c=getchar())!='\n'&&i<KK){if(c>='0'&&c<='9'||c>='A'&&c<='F'||c>='a'&&c<='f')str[i++]=c;else f1=1;}str[i]='\0';if(f1==1)printf("Data Error!\n");else printf("result is :%d\n",sjz(str));}long sjz(char a[]){long n=0,i;;for(i=0;a[i]!='\0';i++){if(a[i]>='0'&&a[i]<='9')n=n*16+a[i]-'0';if(a[i]>='A'&&a[i]<='F')n=n*16+a[i]-'A'+10;if(a[i]>='a'&&a[i]<='f')n=n*16+a[i]-'a'+10;}return n;}习题八(P181) （注：无答案的题不在本课程讲授范围内）一．单选题1.D2.A3.C4.D5.C6.B7.C8.C9.C 13.D 14.B 15.A 16.C 17.A 二．填空题1. 指向取地址2. 2 +23. 286. *(p+5)7. ABCD A三．阅读程序题1．102．103．04. 3,65. 1 2 3 4四．程序填空题1．a,b,c或者&x,&y,&z max=*b或者max=y max=*c或者max=z2. ++ =*q ++ ++4. int *a,int *b b[j]=a[i] b[i++]5. *str+=3 *str>’z’&&*str<’a’||*str>’z’ a6. else 0 t[2*j+1]五．程序设计题1.#include <stdio.h>main(){int a[10],*p,*q,t;p=a;for(q=a;q<a+10;q++)scanf("%d",q);for(q=a;q<a+10;q++)printf("%5d",*q);printf("\n");q--;for(;p<q;p++,q--){t=*p;*p=*q;*q=t;}for(p=a;p<a+10;p++)printf("%5d",*p);printf("\n");}2.#include <stdio.h>#include <string.h>void main( ){ char a[50],*p1,*p2,t,n;gets(a);n=strlen(a);p1=a; p2=a+n-1;for(; p1<p2; p1++, p2--){t=*p1; *p1=*p2; *p2=t;}puts(a);}另一做法#include <stdio.h>main(){char str[20],*p=str;gets(str);while(*p)p++;p--;while(p>=str){printf("%c",*p);p--;}printf("\n");}3.#include <stdio.h>main(){int a[10],*p,*max,*min,t;for(p=a;p<a+10;p++)scanf("%d",p); for(p=a;p<a+10;p++)printf("%5d",*p); printf("\n");max=min=a;for(p=a+1;p<a+10;p++){if(*p>*max)max=p;if(*p<*min)min=p;}t=a[0];a[0]=*min;*min=t;t=a[9];a[9]=*max;*max=t;for(p=a;p<a+10;p++)printf("%5d",*p); printf("\n");}5.#include <stdio.h>int length(char *s){int n=0;while(*s){n++;s++;}return n;}main(){char str[20];int n;gets(str);n=length(str);printf("The string length is %d\n",n);}6.#include <stdio.h>main(){char str[81],*p=str,*q,t;gets(str);printf("The origenal string:\n");puts(str);for(p=str;*(p+1);p++)for(q=p+1;*q;q++)if(*q<*p){t=*p;*p=*q;*q=t;}printf("The result string:\n");puts(str);}习题九(P222) （注：无答案的题不在本课程讲授范围内）一．单选题1.D2.A3.B4.D 8.B 9.B 10.C二．填空题1. 结构体成员结构体指针指向2. 343. 224．ex三．阅读程序题1．92．10,x3．134. 46 40 415. 06. 3839。

《C语言程序设计能力教程（第二版）》课后作业及实训题参考答案第1章进入C语言程序世界二、1. I love China!printf("we are students.\n")2. 6项目实训题参考答案1．编写一个C程序，输出以下信息：* * * * * * * * * * * * * * * * * * * *I am a student!* * * * * * * * * * * * * * * * * * * *main(){ printf("********************\n");printf(" I am a student!\n ");printf("********************\n");}2．已知立方体的长、宽、高分别是10cm、20cm、15cm，编写程序，求立方体体积。

解：main(){int a,b,c,v;a=10;b=20;c=15;v=a*b*c;printf("v=%d",v);}本程序运行结果为:v=3000第2章编制C程序的基础知识一选择题C B A B A C C二操作题，2，-8，23.000000，2.500000，-8.0000002. ABC DEFGHwhy is 21+35 equal 523.3 14 32 31 24. aa bb cc abcA N项目实训题1．定义一个符号常量M为5和一个变量n值为2，把它们的乘积输出。

#define M 5main(){ int n,c;n=2; c=M*n;printf("%d\n",c); }2．编程求下面算术表达式的值。

（1）x+a%3*(int)(x+y)%2/4，设x=2.5,a=7,y=4.7;（2）(float)(a+b)/2+(int)x%(int)y，设a=2,b=3,x=3.5,y=2.5。

c语言程序设计现代方法（第二版）习题答案（5篇）第一篇：c语言程序设计现代方法(第二版)习题答案Chapter 2 Answers to Selected Exercises 2.[was #2](a)The program contains one directive(#include)and four statements(three calls of printf and one return).(b)Parkinson's Law: Work expands so as to fill the time available for its completion.3.[was #4] #includeint main(void){ int height = 8, length = 12, width = 10, volume;volume = height * length * width;printf(“Dimensions: %dx%dx%dn”, length, width, height);printf(“Volume(cubic inches): %dn”, volume);printf(“Dimensional weight(pounds): %dn”,(volume + 165)/ 166);return 0;} 4.[was #6] Here's one possible program: #include int main(void){ int i, j, k;float x, y, z;printf(“Value of i: %dn”, i);printf(“Value of j: %dn”, j);printf(“Value of k: %dn”, k);printf(“Value of x: %gn”, x);printf(“Value of y: %gn”, y);printf(“Value of z: %gn”, z);return 0;} When compiled using GCC and then executed, this program produced the following output: Value of i: 5618848 Value of j: 0 Value of k: 6844404 Value of x: 3.98979e-34 Value of y: 9.59105e-39 Value of z: 9.59105e-39 The values printed depend on many factors, so the chance that you'll get exactly these numbers is small.5.[was #10](a)is not legal because 100_bottles begins with a digit.8.[was #12] There are 14 tokens: a, =,(, 3, *, q,-, p, *, p,), /, 3, and;.Answers to Selected Programming Projects 4.[was #8;modified] #includeint main(void){ float original_amount, amount_with_tax;printf(“Enter an amount: ”);scanf(“%f”, &original_amount);amount_with_tax = original_amount * 1.05f;printf(“With tax added: $%.2fn”, amount_with_tax);return 0;} The amount_with_tax variable is unnecessary.If we remove it, the program is slightly shorter: #includeint main(void){ float original_amount;printf(“Enter an amount: ”);scanf(“%f”, &original_amount);printf(“With tax added: $%.2fn”, original_amount * 1.05f);return 0;}Chapter 3 Answers to Selected Exercises 2.[was #2](a)printf(“%-8.1e”, x);(b)printf(“%10.6e”, x);(c)printf(“%-8.3f”, x);(d)printf(“%6.0f”, x);5.[was #8] The values of x, i, and y will be 12.3, 45, and.6, respectively.Answers to Selected Programming Projects 1.[was #4;modified] #includeint main(void){ int month, day, year;printf(“Enter a date(mm/dd/yyyy): ”);scanf(“%d/%d/%d”, &month, &day, &year);printf(“You entered the date %d%.2d%.2dn”, year, month, day);return 0;} 3.[was #6;modified] #includeint main(void){ int prefix, group, publisher, item, check_digit;printf(“Enter ISBN: ”);scanf(“%d-%d-%d-%d-%d”, &prefix, &group, &publisher, &item, &check_digit);printf(“GS1 prefix: %dn”, prefix);printf(“Group identifier: %dn”, group);printf(“Publisher code: %dn”, publisher);printf(“Item number: %dn”, item);printf(“Check digit: %dn”, check_digit);/* The five printf calls can be combined as follows:printf(“GS1 prefix: %dnGroup identifier: %dnPublishercode: %dnItem number: %dnCheck digit: %dn”, prefix, group, publisher, item, check_digit);*/return 0;}Chapter 4 Answers to Selected Exercises 2.[was #2] Not in C89.Suppose that i is 9 and j is 7.The value of(-i)/j could be either –1 or –2, depending on the implementation.On the other hand, the value of-(i/j)is always –1, regardless of the implementation.In C99, on the other hand, the value of(-i)/j must be equal to the value of-(i/j).9.[was #6](a)63 8(b)3 2 1(c)2-1 3(d)0 0 0 13.[was #8] The expression ++i is equivalent to(i += 1).The value of both expressions is i after the increment has been performed.Answers to Selected Programming Projects 2.[was #4] #include int main(void){ int n;printf(“Enter a three-digit number: ”);scanf(“%d”, &n);printf(“The reversal is: %d%d%dn”, n % 10,(n / 10)% 10, n / 100);return 0;}Chapter 5 Answers to Selected Exercises 2.[was #2](a)1(b)1(c)1(d)1 4.[was #4](i > j)12, minutes);return 0;} 4.[was #8;modified] #includeint main(void){ int speed;printf(“Enter a wind speed in knots: ”);scanf(“%d”, &speed);if(speed < 1)printf(“Calmn”);else if(speed <= 3)printf(“Light airn”);else if(speed <= 27)printf(“Breezen”);else if(speed <= 47)printf(“Galen”);else if(speed <= 63)printf(“Stormn”);else printf(“Hurricanen”);return 0;} 6.[was #10] #includeint main(void){ int check_digit, d, i1, i2, i3, i4, i5, j1, j2, j3, j4, j5, first_sum, second_sum, total;printf(“Enter the first(single)digit: ”);scanf(“%1d”, &d);printf(“Enter first group of five digits: ”);scanf(“%1d%1d%1d%1d%1d”, &i1, &i2, &i3, &i4, &i5);printf(“Enter second group of five digits: ”);scanf(“%1d%1d%1d%1d%1d”, &j1, &j2, &j3, &j4, &j5);printf(“Enter the last(single)digit: ”);scanf(“%1d”, &check_digit);first_sum = d + i2 + i4 + j1 + j3 + j5;second_sum = i1 + i3 + i5 + j2 + j4;total = 3 * first_sum + second_sum;if(check_digit == 91)% 10))printf(“VALIDn”);else printf(“NOT VALIDn”);return 0;} 10.[was #14] #includeint main(void){ int grade;printf(“Enter numerical grade: ”);scanf(“%d”, &grade);if(grade < 0 || grade > 100){ printf(“Illegal graden”);return 0;}switch(grade / 10){ case 10: case 9: printf(“Letter grade: An”);break;case 8: printf(“Letter grade: Bn”);break;case 7: printf(“Letter grade: Cn”);break;case 6: printf(“Letter grade: Dn”);break;case 5: case 4: case 3: case 2: case 1: case 0: printf(“Letter grade: Fn”);break;}return 0;}Chapter 6 Answers to Selected Exercises 4.[was #10](c)is not equivalent to(a)and(b), because i is incremented before the loop body is executed.10.[was #12] Consider the following while loop: while(…){…continue;… } The equivalent code using goto would have the following appearance: while(…){…goto loop_end;…loop_end:;/* null statement */ } 12.[was #14] for(d = 2;d * d <= n;d++)if(n % d == 0)break;The if statement that follows the loop will need to be modified as well: if(d * d <= n)printf(“%d is divisible by %dn”, n, d);else printf(“%d is primen”, n);14.[was #16] The problem is the semicolon at the end of the first line.If we remove it, the statement is now correct: if(n % 2 == 0)printf(“n is evenn”);Answers to Selected Programming Projects 2.[was #2] #includeint main(void){ int m, n, remainder;printf(“Enter two integers: ”);scanf(“%d%d”, &m, &n);while(n!= 0){ remainder = m % n;m = n;n = remainder;} printf(“Greatest common divisor: %dn”, m);return 0;} 4.[was #4] #includeint main(void){ float commission, value;printf(“Enter value of trade: ”);scanf(“%f”, &value);while(value!= 0.0f){ if(value < 2500.00f)commission = 30.00f +.017f * value;else if(value < 6250.00f)commission = 56.00f +.0066f * value;else if(value < 20000.00f)commission = 76.00f +.0034f * value;else if(value < 50000.00f)commission = 100.00f +.0022f * value;else if(value < 500000.00f)commission = 155.00f +.0011f * value;else commission = 255.00f +.0009f * value;if(commission < 39.00f)commission = 39.00f;printf(“Commission: $%.2fnn”, commission);printf(“Enter value of trade: ”);scanf(“%f”, &value);}return 0;} 6.[was #6] #includeint main(void){ int i, n;printf(“Enter limit on maximum square: ”);scanf(“%d”, &n);for(i = 2;i * i <= n;i += 2)printf(“%dn”, i * i);return 0;} 8.[was #8] #includeint main(void){ int i, n, start_day;printf(“Enter number of days in month: ”);scanf(“%d”, &n);printf(“Enter starting day of the week(1=Sun, 7=Sat): ”);scanf(“%d”, &start_day);/* print any leading “blank dates” */ for(i = 1;i < start_day;i++)printf(“ ”);/* now print the calendar */ for(i = 1;i <= n;i++){ printf(“%3d”, i);if((start_day + i1 && y >= 0 && y <= npass;card++){ rank = hand[card][RANK];suit = hand[card][SUIT];if(hand[card+1][RANK] < rank){ hand[card][RANK] = hand[card+1][RANK];hand[card][SUIT] = hand[card+1][SUIT];hand[card+1][RANK] = rank;hand[card+1][SUIT] = suit;} }/* check for flush */ suit = hand[0][SUIT];for(card = 1;card < NUM_CARDS;card++)if(hand[card][SUIT]!= suit)flush = false;/* check for straight */ for(card = 0;card < NUM_CARDS1 && num_in_rank[0] > 0 && num_in_rank[NUM_RANKS-1] > 0){ straight = true;return;}/* check for 4-of-a-kind, 3-of-a-kind, and pairs */ for(rank = 0;rank < NUM_RANKS;rank++){ if(num_in_rank[rank] == 4)four = true;if(num_in_rank[rank] == 3)three = true;if(num_in_rank[rank] == 2)pairs++;} }/********************************************************** * print_result: Prints the classification of the hand, * * based on the values of the external * * variables straight, flush, four, three, * * and pairs.* **********************************************************/ void print_result(void){ if(straight && flush)printf(“Straight flush”);else if(four)printf(“Four of a kind”);else if(three &&pairs == 1)printf(“Full house”);else if(flush)printf(“Flush”);else if(straight)printf(“Straight”);else if(three)printf(“Three of a kind”);else if(pairs == 2)printf(“Two pairs”);else if(pairs == 1)printf(“Pair”);else printf(“High card”);printf(“nn”);}Chapter 11 Answers to Selected Exercises 2.[was #2](e),(f), and(i)are legal.(a)is illegal because p is a pointer to an integer and i is an integer.(b)is illegal because *p is an integer and &i is a pointer to an integer.(c)is illegal because &p is a pointer to a pointer to an integer and q is a pointer to an integer.(d)is illegal for reasons similar to(c).(g)is illegal because p is a pointer to an integer and *q is an integer.(h)is illegal because *p is an integer and q is a pointer to an integer.4.[was #4;modified] void swap(int *p, int *q){ int temp;temp = *p;*p = *q;*q = temp;} 6.[was #6] void find_two_largest(int a[], int n, int *largest, int *second_largest){ int i;if(a[0] > a[1]){ *largest = a[0];*second_largest = a[1];} else { *largest = a[1];*second_largest = a[0];}for(i = 2;i < n;i++)if(a[i] > *largest){ *second_largest = *largest;*largest = a[i];} else if(a[i] > *second_largest)*second_largest = a[i];}Chapter 12 Answers to Selected Exercises 2.[was #2] The statement is illegal because pointers cannot be added.Here's a legal statement that has the desired effect: middle = low +(highlow)/ 2 is an integer, not a pointer, so it can legally be added to low.4.[was #6] int *top_ptr;void make_empty(void){ top_ptr = &contents[0];}bool is_empty(void){ return top_ptr == &contents[0];}bool is_full(void){ return top_ptr == &contents[STACK_SIZE];}6.[was #10;modified] int sum_array(const int a[], int n){ int *p, sum;sum = 0;for(p = a;p < a + n;p++)sum += *p;return sum;} 13.[was #12;modified] #define N 10double ident[N][N], *p;int num_zeros = N;for(p = &ident[0][0];p <= &ident[N-1][N-1];p++)if(num_zeros == N){ *p = 1.0;num_zeros = 0;} else { *p = 0.0;num_zeros++;} 15.[was #14] int *p;for(p = temperatures[i];p < temperatures[i] + 24;p++)printf(“%d ”, *p);Answers to Selected Programming Projects 1.[was #4](a)#include#define MSG_LEN 80 /* maximum length of message */int main(void){ char msg[MSG_LEN];int i;printf(“Enter a message: ”);for(i = 0;i < MSG_LEN;i++){ msg[i] = getchar();if(msg[i] == 'n')break;} printf(“Reversal is: ”);for(i--;i >= 0;i--)putchar(msg[i]);putchar('n');return 0;}(b)#include#define MSG_LEN 80 /* maximum length of message */int main(void){ char msg[MSG_LEN], *p;printf(“Enter a message: ”);for(p = &msg[0];p < &msg[MSG_LEN];p++){ *p = getchar();if(*p == 'n')break;} printf(“Reversal is: ”);for(p--;p >= &msg[0];p--)putchar(*p);putchar('n');return 0;} 3.[was #8] #include#define MSG_LEN 80 /* maximum length of message */int main(void){ char msg[MSG_LEN], *p;printf(“Enter a message: ”);for(p = msg;p < msg + MSG_LEN;p++){ *p = getchar();if(*p == 'n')break;}printf(“Reversal is: ”);for(p--;p >= msg;p--)putchar(*p);putchar('n');return 0;}Chapter 13 Answers to Selected Exercises 2.[was #2](a)Illegal;p is not a character.(b)Legal;output is a.(c)Legal;output is abc.(d)Illegal;*p is not a pointer.4.[was #4](a)int read_line(char str[], int n){ int ch, i = 0;while((ch = getchar())!= 'n')if(i == 0 && isspace(ch));/* ignore */ else if(i < n)str[i++] = ch;str[i] = '';return i;}(b)int read_line(char str[], int n){ int ch, i = 0;while(!isspace(ch = getchar()))if(i < n)str[i++] = ch;str[i] = '';return i;}(c)int read_line(char str[], int n){ int ch, i = 0;do { ch = getchar();if(i < n)str[i++] = ch;} while(ch!= 'n');str[i] = '';return i;}(d)int read_line(char str[], int n){ int ch, i;for(i = 0;i < n;i++){ ch = getchar();if(ch == 'n')break;str[i] = ch;} str[i] = '';return i;} 6.[was #6] void censor(char s[]){ int i;for(i = 0;s[i]!= '';i++)if(s[i] == 'f' && s[i+1] == 'o' && s[i+2] =='o')s[i] = s[i+1] = s[i+2] = 'x';} Note that the short-circuit evaluation of && prevents the if statement from testing characters that follow the null character.8.[was #10] tired-or-wired? 10.[was #12] The value of q is undefined, so the call of strcpy attempts to copy the string pointed to by p into some unknown area of memory.Exercise 2 in Chapter 17 discusses how to write this function correctly.15.[was #8](a)3(b)0(c)The length of the longest prefix of the string s that consists entirely of characters from the string t.Or, equivalently, the position of the first character in s that is not also in t.16.[was #16] int count_spaces(const char *s){ int count = 0;while(*s)if(*s++ == ' ')count++;return count;} Answers to Selected Programming Projects 1.[was #14] #include #include #define WORD_LEN 20void read_line(char str[], int n);int main(void){ char smallest_word[WORD_LEN+1], largest_word[WORD_LEN+1], current_word[WORD_LEN+1];printf(“Enter word: ”);read_line(current_word, WORD_LEN);strcpy(smallest_word, strcpy(largest_word, current_word));while(strlen(current_word)!= 4){ printf(“Enter word: ”);read_line(current_word, WORD_LEN);if(strcmp(current_word, smallest_word)< 0)strcpy(smallest_word, current_word);if(strcmp(current_word, largest_word)> 0)strcpy(largest_word, current_word);} printf(“nSmallest word: %sn”, smallest_word);printf(“Largest word:%sn”, largest_word);return 0;}void read_line(char str[], int n){ int ch, i = 0;while((ch = getchar())!= 'n')if(i < n)str[i++] = ch;str[i] = '';} 4.[was #18] #includeint main(int argc, char *argv[]){ int i;for(i = argc-1;i > 0;i--)printf(“%s ”, argv[i]);printf(“n”);return 0;} 6.[was #20] #include #include #include#define NUM_PLANETS 9int string_equal(const char *s, const char *t);int main(int argc, char *argv[]){ char *planets[] = {“Mercury”, “Venus”, “Earth”, “Mars”, “Jupiter”, “Saturn”, “Uranus”, “Neptune”, “Pluto”};int i, j;for(i = 1;i < argc;i++){ for(j = 0;j < NUM_PLANETS;j++)if(string_equal(argv[i],planets[j])){ printf(“%s is planet %dn”, argv[i], j + 1);break;} if(j == NUM_PLANETS)printf(“%s is not a planetn”, argv[i]);} return 0;}int string_equal(const char *s, const char *t){ int i;for(i = 0;toupper(s[i])== toupper(t[i]);i++)if(s[i] == '')return 1;return 0;}Chapter 14 Answers to Selected Exercises 2.[was #2] #define NELEMS(a)((int)(sizeof(a)/ sizeof(a[0])))4.[was #4](a)One problem stems from the lack of parentheses around the replacement list.For example, the statement a = 1/AVG(b, c);will be replaced by a = 1/(b+c)/2;Even if we add the missing parentheses, though, the macro still has problems, because it needs parentheses around x and y in the replacement list.The preprocessor will turn the statement a = AVG(bd);into a =((bd)/2);which is equivalent to a =((bd)/2);Here's the final(corrected)version of the macro: #define AVG(x,y)(((x)+(y))/2)(b)The problem is the lack of parentheses around the replacement list.For example, a = 1/AREA(b, c);becomes a = 1/(b)*(c);Here's the corrected macro: #define AREA(x,y)((x)*(y))5.[was #6](a)The call of putchar expands into the following statement: putchar(('a'<=(s[++i])&&(s[++i])<='z'?(s[++i])-'a'+'A':(s[++i])));The character a is less than or equal to s[1](which is b), yielding a true condition.The character s[2](which is c)is less than or equal to z, which is also true.The value printed is s[3]-'a'+'A', which is D(assuming that the character set is ASCII).(b)The character a is not less than or equal to s[1](which is 1)so the test condition is false.The value printed is s[2], which is 2.7.[was #8](a)long long_max(long x, long y){ return x > y ? x : y;} The preprocessor would actually put all the tokens on one line, but this version is more readable.(b)The problem with types such as unsigned long is that they require two words, which prevents GENERIC_MAX from creating the desired function name.For example, GENERIC_MAX(unsigned long)would expand intounsigned long unsigned long_max(unsigned long x, unsigned long y){ return x > y ? x : y;}(c)To make GENERIC_MAX work with any basic type, use a type definition to rename the type: typedef unsigned long ULONG;We can now write GENERIC_MAX(ULONG).12.[was #10](c)and(e)will fail, since M is defined.14.[was #12;modified] Here's what the program will look like after preprocessing: Blank line Blank line Blank line Blank line Blank line Blank line Blank lineint main(void){ int a[= 10], i, j, k, m;Blank line i = j;Blank line Blank line Blank linei = 10 * j+1;i =(x,y)x-y(j, k);i =((((j)*(j)))*(((j)*(j))));i =(((j)*(j))*(j));i = jk;puts(“i” “j”);Blank line i = SQR(j);Blank line i =(j);return 0;} Some preprocessors delete white-space characters at the beginning of a line, so your results may vary.Three lines will cause errors when the program is compiled.Two contain syntax errors: int a[= 10], i, j, k, m;i =(x,y)x-y(j, k);The third refers to an undefined variable: i = jk;Chapter 15 Answers to Selected Exercises 2.[was #2](b).Function definitions should not be put in a header file.If a function definition appears in a header file that is included by two(or more)source files, the program can't be linked, since the linker will see two copies of the function.6.[was #8](a)main.c, f1.c, and f2.c.(b)f1.c(assuming that f1.h is not affected by the change).(c)main.c, f1.c, and f2.c, since all three include f1.h.(d)f1.c and f2.c, since both include f2.h.Chapter 16 Answers to Selected Exercises 2.[was #2;modified](a)struct { double real, imaginary;} c1, c2, c3;(b)struct { double real, imaginary;} c1 = {0.0, 1.0}, c2 = {1.0, 0.0}, c3;(c)Only one statement is necessary: c1 = c2;(d)c3.real = c1.real + c2.real;c3.imaginary = c1.imaginary +c2.imaginary;4.[was #4;modified](a)typedef struct { double real, imaginary;} Complex;(b)Complex c1, c2, c3;(c)Complex make_complex(double real, double imaginary){ Complex c;c.real = real;c.imaginary = imaginary;return c;}(d)Complex add_complex(Complex c1, Complex c2){ Complex c3;c3.real = c1.real + c2.real;c3.imaginary = c1.imaginary + c2.imaginary;return c3;} 11.[was #10;modified] The a member will occupy 8 bytes, the union e will take 8 bytes(the largest member, c, is 8 bytes long), and the array f will require 4 bytes, so the total space allocated for s will be 20 bytes.14.[was #12;modified](a)double area(struct shape s){ if(s.shape_kind == RECTANGLE)return s.u.rectangle.height * s.u.rectangle.width;else return 3.14159 * s.u.circle.radius * s.u.circle.radius;}(b)struct shape move(struct shape s, int x, int y){ struct shape new_shape = s;new_shape.center.x += x;new_shape.center.y += y;return new_shape;}(c)struct shape scale(struct shape s, double c){ struct shape new_shape = s;if(new_shape.shape_kind == RECTANGLE){ new_shape.u.rectangle.height *= c;new_shape.u.rectangle.width *= c;} else new_shape.u.circle.radius *= c;return new_shape;} 15.[was #14](a)enum week_days {MON, TUE, WED, THU, FRI, SAT, SUN};(b)typedef enum {MON, TUE, WED, THU, FRI, SAT, SUN} Week_days;17.[was #16] All the statements are legal, since C allows integers and enumeration values to be mixed without restriction.Only(a),(d), and(e)are safe.(b)is not meaningful if i has a value other than 0 or 1.(c)will not yield a meaningful result if b has the value 1.Answers to Selected Programming Projects 1.[was #6;modified] #include#define COUNTRY_COUNT((int)(sizeof(country_codes)/sizeof(country_codes[0])))struct dialing_code { char *country;int code;};const struct dialing_code country_codes[] = {{“Argentina”, 54}, {“Bangladesh”, 880}, {“Brazil”, 55}, {“Burma(Myanmar)”, 95}, {“China”, 86}, {“Colombia”, 57}, {“Congo, Dem.Rep.of”, 243}, {“Egypt”, 20}, {“Ethiopia”, 251}, {“France”, 33}, {“Germany”, 49}, {“India”, 91}, {“Indonesia”, 62}, {“Iran”, 98}, {“Italy”, 39}, {“Japan”, 81}, {“Mexico”, 52}, {“Nigeria”, 234}, {“Pakistan”, 92}, {“Philippines”, 63}, {“Poland”, 48}, {“Russia”, 7}, {“South Africa”, 27}, {“South Korea”, 82}, {“Spain”, 34}, {“Sudan”, 249}, {“Thailand”, 66}, {“Turkey”, 90}, {“Ukraine”, 380}, {“United Kingdom”, 44}, {“United States”, 1}, {“Vietnam”, 84}};int main(void){ int code, i;printf(“Enter dialing code: ”);scanf(“%d”, &code);for(i = 0;i < COUNTRY_COUNT;i++)if(code == country_codes[i].code){ printf(“The country with dialing code %d is %sn”, code, country_codes[i].country);re turn 0;} printf(“No corresponding country foundn”);return 0;} 3.[was #8] #include #include “readline.h”#define NAME_LEN 25 #define MAX_PARTS 100struct part { int number;char name[NAME_LEN+1];int on_hand;};int find_part(int number, const struct part inv[], int np);void insert(struct part inv[], int *np);void search(const struct part inv[], int np);void update(struct part inv[], int np);void print(const struct part inv[], int np);/********************************************************** * main: Prompts the user to enter an operation code, * * then calls a function to perform the requested * * action.Repeats until theuser enters the * * command 'q'.Prints an error message if the user * * enters an illegal code.* **********************************************************/ int main(void){ char code;struct part inventory[MAX_PARTS];int num_parts = 0;for(;;){ printf(“Enter operation code: ”);scanf(“ %c”, &code);while(getchar()!= 'n')/* skips to end of line */;switch(code){ case 'i': insert(inventory, &num_parts);break;case 's': search(inventory, num_parts);break;case 'u': update(inventory, num_parts);break;case 'p': print(inventory, num_parts);break;case 'q': return 0;default: printf(“Illegal coden”);} printf(“n”);} } /********************************************************** * find_part: Looks up a part number in the inv array.* * Returns the array index if the part number * * is found;otherwise, returns-1.* **********************************************************/ int find_part(int number, const struct part inv[], int np){ int i;for(i = 0;i < np;i++)if(inv[i].number == number)return i;return-1;}/********************************************************** * insert: Prompts the user for information about a new * * part and then inserts the part into the inv * * array.Prints an error message and returns * * prematurely if the part already exists or the * * array is full.* **********************************************************/ void insert(struct part inv[], int *np){ int part_number;if(*np == MAX_P ARTS){ printf(“Database is full;can't add more parts.n”);return;}printf(“Enter part number: ”);scanf(“%d”, &part_number);if(find_part(part_number, inv, *np)>= 0){ printf(“Part already exists.n”);return;}inv[*np].number = part_number;printf(“Enter part na me: ”);read_line(inv[*np].name, NAME_LEN);printf(“Enter quantity on hand: ”);scanf(“%d”, &inv[*np].on_hand);(*np)++;} /********************************************************** * search: Prompts the user to enter a part number, then * * looks up the part in the inv array.If the * * part exists, prints the name and quantity on * * hand;if not, prints an error message.* **********************************************************/ void search(const struct part inv[], int np){ int i, number;printf(“Enter part number: ”);scanf(“%d”, &number);i = find_part(number, inv, np);if(i >= 0){ printf(“Part name: %sn”, inv[i].name);printf(“Quantity on hand: %dn”, inv[i].on_hand);} else printf(“Part not found.n”);}/********************************************************** * update: Prompts the user to enter a part number.* * Prints an error message if the part can't be * * found in the inv array;otherwise, prompts the * * user to enter change in quantity on hand and * * updates the array.* **********************************************************/ void update(struct part inv[], int np){ int i, number, change;printf(“Enter part number: ”);scanf(“%d”, &number);i = find_part(number, inv, np);if(i >= 0){ printf(“Enter change in quantity on hand: ”);scanf(“%d”, &change);inv[i].on_hand += change;} else printf(“Part not found.n”);}/********************************************************** * print: Prints a listing of all parts in the inv array, * * showing the part number, part name, and * * quantity on hand.Parts are printed in the * * order in which they were entered into the * * array.* **********************************************************/ void print(const struct part inv[], int np){ int i;printf(“Part Number Part Name ” “Quantity on Handn”);for(i = 0;i < np;i++)printf(“%7d %-25s%11dn”, inv[i].number, inv[i].name, inv[i].on_hand);}Chapter 17 Answers to Selected Exercises 2.[was #2;modified] char *duplicate(const char *s){ char *temp = malloc(strlen(s)+ 1);if(temp == NULL)return NULL;strcpy(temp, s);return temp;} 5.[was #6](b)and(c)are legal.(a)is illegal because it tries to reference a member of d without mentioning d.(d)is illegal because it uses-> instead of.to reference the c member of d.7.[was #8] The first call of free will release the space for the first node in the list, making p a dangling pointer.Executing p = p->next to advance to the next node will have an undefined effect.Here's a correct way to write the loop, using a temporary pointer that points to the node being deleted: struct node *temp;p = first;while(p!= NULL){ temp = p;p = p->next;free(temp);} 8.[was #10;modified] #include /* C99 only */ #include #include #include “stack.h”struct node { int value;struct node *next;};struct node *top = NULL;void make_empty(void){ struct node *temp;第二篇：《C语言程序设计教程(第二版)》习题答案《C语言程序设计教程(第二版)》习题答案说明1.本习题答案是我自己做的，错误和疏漏在所难免。

c程序设计第二版习题答案C程序设计第二版习题答案第一章：C语言概述1. 描述C语言的特点。

- C语言是一种结构化编程语言，以其高效性、灵活性和可移植性而闻名。

它支持过程式编程和低级内存操作，适合编写系统软件、操作系统和嵌入式系统。

2. 解释C语言的编译过程。

- C语言的编译过程通常包括预处理、编译、汇编和链接四个步骤。

预处理阶段处理宏定义和条件编译指令，编译阶段将源代码转换为汇编代码，汇编阶段将汇编代码转换为目标代码，链接阶段将目标代码与库函数链接生成可执行文件。

第二章：数据类型、运算符和表达式1. 列举C语言的基本数据类型。

- C语言的基本数据类型包括整型（int）、字符型（char）、浮点型（float和double）以及枚举型（enum）。

2. 说明赋值运算符的用法。

- 赋值运算符（=）用于将一个值赋给一个变量。

例如，`a = 5;`表示将整数5赋给变量a。

第三章：控制语句1. 描述if语句的语法结构。

- if语句的基本语法是：`if (条件) { 语句; }`。

如果条件为真，则执行花括号内的语句。

2. 解释while循环的工作原理。

- while循环在每次迭代之前都会检查条件是否为真。

如果条件为真，循环体内的语句将被执行，然后再次检查条件。

这个过程会一直重复，直到条件为假。

第四章：函数1. 函数的定义和声明。

- 函数定义是创建函数的主体，包括函数名、参数列表和函数体。

函数声明是告诉编译器函数的原型，包括函数名、返回类型和参数列表。

2. 函数的调用过程。

- 函数调用包括函数名和参数列表。

调用时，实际参数的值将传递给形式参数，函数执行完毕后，返回值（如果有的话）将返回给调用者。

第五章：数组和指针1. 数组的基本概念。

- 数组是一种数据结构，可以存储固定大小的同类型元素的集合。

数组的元素可以通过索引访问。

2. 指针的用途。

- 指针用于存储内存地址，可以直接操作内存。

指针可以指向基本数据类型、数组、结构体等。

C语言程序设计(第二版)习题参考题答案-CAL-FENGHAI.-(YICAI)-Company One1C语言程序设计习题参考答案习题 1一、判断题1．在计算机中，小数点和正负号都有专用部件来保存和表示。

2．二进制是由0和1两个数字组成的进制方式。

3．二进制数的逻辑运算是按位进行的，位与位之间没有进位和借位的关系。

4．在整数的二进制表示方法中，0的原码、反码都有两种形式。

5．有符号数有三种表示法：原码、反码和补码。

6．常用字符的ASCII码值从小到大的排列规律是：空格、阿拉伯数字、大写英文字母、小写英文字母。

解：1．F 2．T 3．T 4．T 5．T 6．T二、单选题1．在计算机中，最适合进行数值加减运算的数值编码是。

A. 原码B. 反码C. 补码D. 移码2．已知英文小写字母m的ASCII码为十进制数109，则英文小写字母y的ASCII码为十进制数。

A. 112B. 120C. 121D. 1223．关于ASCII码，在计算机中的表示方法准确地描述是。

A. 使用8位二进制数，最右边一位为1B. 使用8位二进制数，最左边一位为1C. 使用8位二进制数，最右边一位为0D. 使用8位二进制数，最左边一位为04．设在机器字长4位，X＝0111B，Y＝1011B，则下列逻辑运算中，正确的是___________。

A. X∧Y＝1000B. X∨Y＝1111C. X⊕Y＝0011D. ¯Y＝10005．下列叙述中正确的是（）。

A．高级语言就是机器语言B．汇编语言程序、高级语言程序都是计算机程序，但只有机器语言程序才是计算机可以直接识别并执行的程序C．C语言因为具有汇编语言的一些特性，所以是汇编语言的一种D．C源程序经过编译、连接，若正确，执行后就能得到正确的运行结果6．用C语言编写的源程序经过编译后，若没有产生编译错误，则系统将（）。

A．生成可执行文件B．生成目标文件C．输出运行结果D．自动保存源文件7．下列叙述中不正确的是（）。

c语言程序设计(第二版)习题答案《C语言程序设计(第二版)习题答案》C语言是一种通用的高级程序设计语言，广泛应用于系统软件、应用软件、驱动程序、网络软件、嵌入式软件等领域。

《C语言程序设计(第二版)》是一本经典的教材，对于学习C语言的人来说具有重要的指导意义。

在学习过程中，通过做习题可以更好地巩固所学知识。

本文将为大家介绍一些《C语言程序设计(第二版)》中常见的习题答案，希望对大家的学习有所帮助。

1. 编写一个C程序，实现输入两个整数，输出它们的和。

```c#include <stdio.h>int main() {int num1, num2, sum;printf("请输入两个整数：");scanf("%d %d", &num1, &num2);sum = num1 + num2;printf("它们的和为：%d\n", sum);return 0;}```2. 编写一个C程序，实现输入一个整数，判断它是否为偶数。

```c#include <stdio.h>int main() {int num;printf("请输入一个整数：");scanf("%d", &num);if (num % 2 == 0) {printf("是偶数\n");} else {printf("不是偶数\n");}return 0;}```3. 编写一个C程序，实现输入一个字符，判断它是否为大写字母。

```c#include <stdio.h>int main() {char ch;printf("请输入一个字符：");scanf(" %c", &ch);if (ch >= 'A' && ch <= 'Z') {printf("是大写字母\n");} else {printf("不是大写字母\n");}return 0;}```以上是《C语言程序设计(第二版)》中一些常见习题的答案，希望对大家的学习有所帮助。

C语言程序设计（第2版）课后习题答案第一章1.请参照本章例题，编写一个C程序，输出以下信息：**************************V ery good!**************************解：#include<stdio.h>void main(){printf(“**************************”);printf(“\n”);printf(“V ery good!\n”);printf(“\n”);printf(“**************************”);}2.编写一个C程序，输入a、b、c三个值，输出其中最大值。

解：#include<stdio.h>void main(){int a,b,c,max;printf(“请输入三个数a,b,c:\n”);scanf(“%d,%d,%d”,&a,&b,&c);max=a;if(max<b) max=b;if(max<c) max=c;printf(“最大数为: %d”,max);}第二章1.假如我国国民生产总值的年增长率为10%，计算10年后我国国民生产总值与现在相比增长多少百分比。

计算公式为P=(1+r)^n，r为年增长率；n为年数；P为与现在相比的百分比。

解：#include<stdio.h>#include<math.h>void main(){double P, r=0.1, n=10;P=pow((1+r), n);printf(“%lf\n”, P);}3.请编程序将“China”译成密码，译码规律是用原来字母后面的第4个字母代替原来的字母。

例如，字母“A”后面第4个字母是“E”，“E”代替“A”。

因此，“China”应译为“Glmre”。

请编一程序，用赋初值的方法使cl、c2、c3、c4、c5五个变量的值分别为‟C‟、‟h‟、‟i‟、‟n‟、‟a‟，经过运算，使c1、c2、c3、c4、c5分别变为‟G‟、‟l‟、‟m‟、‟r‟、‟e‟，并输出。

c 语言程序设计教程第二版课后习题答案C语言程序设计教程第二版课后习题答案在学习C语言程序设计的过程中，课后习题是巩固知识、提高编程能力的重要环节。

本文将为大家总结C语言程序设计教程第二版课后习题的答案，希望对大家的学习有所帮助。

第一章课后习题答案1.1 编写一个C程序，输出"Hello, World!"。

```c#include <stdio.h>int main() {printf("Hello, World!\n");return 0;}```1.2 编写一个C程序，计算并输出5的阶乘。

```c#include <stdio.h>int main() {int i, fact = 1;for (i = 1; i <= 5; i++) {fact *= i;}printf("5的阶乘为%d\n", fact);return 0;}```1.3 编写一个C程序，输入一个整数n，计算并输出1到n的和。

```c#include <stdio.h>int main() {int n, sum = 0, i;printf("请输入一个整数：");scanf("%d", &n);for (i = 1; i <= n; i++) {sum += i;}printf("1到%d的和为%d\n", n, sum);return 0;}```第二章课后习题答案2.1 编写一个C程序，输入一个整数n，计算并输出n的平方。

```c#include <stdio.h>int main() {int n;printf("请输入一个整数：");scanf("%d", &n);printf("%d的平方为%d\n", n, n * n);return 0;}```2.2 编写一个C程序，输入一个整数n，判断并输出n是奇数还是偶数。

WORD格式可编辑C语言程序设计习题参考答案习题 1一、判断题1．在计算机中，小数点和正负号都有专用部件来保存和表示。

2．二进制是由0和1两个数字组成的进制方式。

3．二进制数的逻辑运算是按位进行的，位与位之间没有进位和借位的关系。

4．在整数的二进制表示方法中，0的原码、反码都有两种形式。

5．有符号数有三种表示法：原码、反码和补码。

6．常用字符的ASCII码值从小到大的排列规律是：空格、阿拉伯数字、大写英文字母、小写英文字母。

解：1．F 2．T 3．T 4．T 5．T 6．T二、单选题1．在计算机中，最适合进行数值加减运算的数值编码是。

A. 原码B. 反码C. 补码D. 移码2．已知英文小写字母m的ASCII码为十进制数109，则英文小写字母y的ASCII码为十进制数。

A. 112B. 120C. 121D. 1223．关于ASCII码，在计算机中的表示方法准确地描述是。

A. 使用8位二进制数，最右边一位为1B. 使用8位二进制数，最左边一位为1C. 使用8位二进制数，最右边一位为0D. 使用8位二进制数，最左边一位为04．设在机器字长4位，X＝0111B，Y＝1011B，则下列逻辑运算中，正确的是___________。

A. X∧Y＝1000B. X∨Y＝1111C. X⊕Y＝0011D. ¯Y＝10005．下列叙述中正确的是（）。

A．高级语言就是机器语言B．汇编语言程序、高级语言程序都是计算机程序，但只有机器语言程序才是计算机可以直接识别并执行的程序C．C语言因为具有汇编语言的一些特性，所以是汇编语言的一种D．C源程序经过编译、连接，若正确，执行后就能得到正确的运行结果6．用C语言编写的源程序经过编译后，若没有产生编译错误，则系统将（）。

A．生成可执行文件B．生成目标文件C．输出运行结果D．自动保存源文件7．下列叙述中不正确的是（）。

A．main函数在C程序中必须有且只有一个B. C程序的执行从main函数开始，所以main函数必须放在程序最前面C. 函数可以带参数，也可以不带参数。

C 语言程序设计习题答案习题一 C 语言程序设计概述一、名词解释（1）程序P1 （2）程序设计P1 （3）机器语言P1 （4）汇编程序P2（5）高级语言P2 （6）编译程序P3 （7）解释程序P3 （8）算法P4（9）结构化的程序设计P9二、简述题1. 设计程序时应遵循哪些基本原则？P4答：正确性、可靠性、简明性、有效性、可维护性、可移植性。

2. 算法的要素是什么？算法具有哪些特点？答：算法的要素是：操作与控制结构；算法的特点有：有穷性、确定性、有效性、有零个或多个输入、有一个或多个输出。

3. 算法的表示形式有哪几种？答：算法的表示形式有：自然语言、传统流程图、伪代码、结构化的流程图（N_S 流程图,盒图）。

4. 有哪三种基本结构？答：三种基本结构是：顺序结构、选择结构和循环结构。

5. 传统流程图与N-S 流程图最大的区别是什么？答：N-S 流程图去掉了在传统流程图中常用的流程线，使得程序的结构显得更加清晰、简单。

三、用传统流程图、N-S 图分别表示求解以下问题的算法。

1. 有3个数a ，b ，c ，要求按由大到小的顺序把它们输出。

2. 依次将10个数输入，求出其中最大的数 和最小的数并输出。

3. 求1+2+3+…+100的值。

4. 求1×2×3×…×10的值。

5. 求下列分段函数的值。

6. 求100~200之间的所有素数。

7. 求一元二次方程ax 2+bx+c=0的根。

分别考虑d=b 2-4ac 大于0、等于0和小于0三种情况。

四、注释下面C 程序的各个组成部分。

main() /*主函数 */{ /*程序开始 */int a,k,m; /*定义三个用来存放整数的变量 */a=10; /*将整数10赋值给变量a */k=2; /*将整数2赋值给变量k */m=1; /*将整数1赋值给变量1 */a=(k+m)*k/(k-m); /*先求出算术表达式的值，并将其赋值给变量a */printf("%d\n",a); /*在屏幕上打印出变量a 的值 */} /*程序结束 */习题二 数据类型、运算符与表达式一、选择题1～10：BCDCB DDBCA11~20: ADDAA DBADC21~28: DABAD CDDY= 3X （X<1） 4X-1 （X=1） 5(X-1)+6 （1<X<5） 6-3X （X ≥5） 输入一个数给x X<=1 Yes no X<1 x<5 Yes no yes no Y=3x y=4x-1 y=5x+1 y=6-3x 输出s 的值 i =100 当i<=200时 n=2; flag=1; 当n< i 时i 能否被n 整除?yes no flag=0 n = n+1 flag=1?yes no输出i 的值i = i+1二、填空题1．字母L 或字母l2．字符或%c 、整数或%d3．在程序运行过程中，其值可以在一定的围变化的量。

c语⾔程序设计教程（第2版）课后题及模拟题参考答案c语⾔程序设计教程（第2版）课后题及模拟题参考答案习题1 (4)1-1 填空题 (4)1-2 思考题 (4)1-3 编程题 (5)习题2 (6)2-1 单选题 (6)2-2 思考题 (6)习题3 (7)3-1 选择题 (7)3-2 填空题 (7)3-3 编程题 (8)习题4 (11)4-1单选题 (11)4-2填空题 (11)4-3 编程题 (11)习题5 (16)5-1单选题 (16)5-2填空题 (16)5-3 编程题 (16)习题6 (22)6-1单选题 (22)6-2填空题 (22)习题7 (25)7-1单选题 (25)7-2填空题 (25)7-3 编程题 (25)习题8 (26)8-1单选题 (26)8-2填空题 (26)8-3 编程题 (26)习题9 (30)9-1单选题 (30)9-2填空题 (30)9-3 编程题 (30)习题10 (38)10-1单选题 (38)10-2填空题 (38)10-3 编程题 (38)习题11 (41)11-1单选题 (41)11-2填空题 (41)习题12 (42)12-1单选题 (42)12-2 填空题 (42)实验1 熟悉Visual C++6.0可视化集成开发环境 (43)实验2 顺序结构程序设计 (43)实验3 选择结构程序设计 (43)实验4 循环结构程序设计 (44)实验5 函数 (47)实验6 数组 (54)实验7 指针 (58)实验8 结构体和共⽤体 (61)实验9 ⽂件 (63)实验10 综合编程 (64)模拟试卷（⼀）参考答案 (65)模拟试卷（⼆）参考答案 (67)习题11-1 填空题1. 函数2. 主函数main(),主函数main()3. 主函数main()4. 函数⾸部，函数体5. {, }6. /*, */7. 顺序结构，选择结构，循环结构8. .c, .obj, .exe1-2 思考题1. 结构化程序设计是指：为使程序具有⼀个合理的结构以保证程序正确性⽽规定的⼀套如何进⾏程序设计的原则。