C语言程序设计答案曹计昌

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1、第一章习题1.4原码：对于一个二进制数X，如果规定其最高位为符号位，其余各位为该数的绝对值，并且规定符号位值为0表示正，为1表示负，采用这种方式的二进制编码称为该二进制数X的原码。补码：正数的补码等于正数的原码，负数的补码为其原码除符号位不动外，其余各位变反再加1所得。反码：对于正数而言，反码与原码相同；对于负数而言，反码符号位的定义与原码相同，但需要将对应原码的数值位按位变反。1.5 和：10101010差：000100001.6 和 01073 差 -03371.7 和 0x1AABA差 -0x53201.8（251）10=（11111011）2=（373）8=（FB）161.10在16位

2、机中，157补= 0000000010011101 -153补157-153=157+(-153)= (0000000010011101) 22=(0000000000000100) 2=(4) 101.14算法设计：用变量s存储累加和，k表示计数描述为：（1）定义变量s，k。（2）s清零，k赋初值1。（3）判断k101？如果是，顺序执行（4）；否则转步骤（5）；（4）k加到累加和变量s中，k加1；转步骤（3）。（5）输出累加和s。（6）结束。开始结束int s=0,k=1;k101?s=s+k;k=k+1;输出sNY1.16 第二章习题2.2(1) x, +, +, y(2)-, 0xabL

3、(3)2.89e+12L(4)”String+” FOO”(5)x, *, *, 2(6)”X?/”(7)a, ?, b(8)x, -, +=, y(9)intx, =, +, 10(10)”String”, “FOO”2.3不是表识符的如下：4th 首字母为数字 sizeof关键字x*y *不是字母、数字、下划线temp-2 -不是字母、数字、下划线isnt 不是字母、数字、下划线enum 关键字2.4合法常数：.12 0.L 1.E-5 3.F 浮点型常量2L 33333 0377UL 0x9cfU 整型常量“a” “” 字符串常量45 a 字符常量非法常数：必须用转义序列0x1ag 十六

4、进制没有gE20 没有尾数部分18 要用八进制数xa 格式错误，可以是xa“34” 需要转义序列” 需要转义序列2.5(1)int a, b=5;(2)double h;(3)int x=2.3; 0.3 会被截取。(4)const long y=1; 必须赋初值(5)float a= 2.5*g; g 没有定义。(6) int a=b=2; 在 turbo C 中编译出错：未定义的符号b在main函数中。2.6（1）4（2）0（3）1（4）6（5）8（6）0（7）3.00（8）1（9）108（10）02.7 答案不确定（1）a=b=c c未定义（2）正确（3）正确（4）正确（5）a*+-b

5、表达式缺值（6）a|bi 运算的操作数必须是整型，而i不是（7）i*j%a %运算的操作数必须是整型，而a不是（8）正确（9）正确（10）int（a+b） 应该改成（int）（a+b）2.9（1）0（2）-2（3）65535（4）5（5）60（6）113（7）-2（8）-1（9）65532（10）32.10unsigned long encrypt(unsigned long x) unsigned long x0,x1,x2,x3,x4,x5,x6,x7; x0=(x & 0x0000000F) 8; x1=(x & 0x000000F0); x2=(x & 0x00000F00) 8; x

6、3=(x & 0x0000F000); x4=(x & 0x000F0000) 24; x7=(x & 0xF0000000); return(x0|x1|x2|x3|x4|x5|x6|x7);2.11#includevoid main()unsigned long in;unsigned long a,b,c,d;scanf(%ld,&in);/in=1563;a=(in&0xff000000)24;b=(in&0x00ff0000)16;c=(in&0x0000ff00)8;d=in&0x000000ff;printf(%d.%d.%d.%d,a,b,c,d);2.15(k 8)& 0xF

7、F00) | (p & 0x00FF)b?ac?a:c:bc?b:c;max=a b ? (a c) ? a : c):(b c) ? b : c);2.17X=yn2.18（c=0 & c=9）? c 0 : c2.19(a % 3 = 0) & (a % 10 = 5) ? a : 0;第三章习题3.1 函数原型是指对函数的名称、返回值类型、参数的数目和参数类型的说明。其规定了调用该函数的语法格式，即调用形式。putchar函数的原型为：int putchar(int c);puts函数的原型为： int puts(const char *s);printf函数的原型为：int print

8、f(const char *format,);getchar函数的原型为:int getchar_r(void);gets函数的原型为:char * gets_r(char *s);scanf函数的原型为: int scanf(const char *format,);3.2 不同点： puts为非格式输出函数，printf为格式输出函数； puts函数的参数类型和数目一定(一个字符串)，printf函数的参数类型和数目不固定； puts函数输出后会自动换行，printf函数没有这一功能。 相同点：二者都向标准设备输出； 二者返回值类型都为int。3.3 x1=-1,177777,ffff,6

9、5535 x2=-3,177775,fffd,65533 y1=123.456703, 123.457,123.457,123.457 (注意对齐) y2=123.449997,1.23450e+02,123.45 x1(%4d)= -13.4 %c;%c;%f;%f;%lu;%d;%d;%d;%f;%Lf3.5 错误,运行提示为divide error 正确,结果为b 正确,结果为 * 正确 正确,但无法正常从结果中退出 正确 正确,结果为82,63 编译错误,提示 cannot modify a const object 正确 正确3.6 -6.70000 -6 177601 123 -2

10、 03.8#includevoid main() char c; c= getchar_r(); if(c=0&c=A&c=a&c=0&c=A&c=F) printf(%dn,c-A+10); else printf(%dn,c-a+10); else putchar(c);3.9#includevoid main() short num,high,low; printf(Please input a short number:n); scanf(%hd,&num); low = 0x00ff & num; high = 0x00ff & (num 8); printf(The high by

11、te is:%cn, high); printf(The low byte is:%cn, low); 3.10#include stdafx.hint main(int argc, char* argv) unsigned short int x; unsigned short int high,low; printf(input a integer:n); scanf(%d,&x); high = (x12)&0x000f; low = (x12)&0xf000; x= x&0x0ff0; x=x|high|low; printf(%dn,x); return 0;3.11#include

12、void main()unsigned short int x,m,n;unsigned short int result;scanf(%hu%hu%hu,&x,&m,&n);result=(x(m-n+1)(15-n+1);printf(%hun,result);3.12#includevoid main() float f,c; scanf(%f,&f); c=(5*(f-32)/9; printf(%.0f(F)=%.2f(C)n,f,c);或者#includevoid main()int f;float c; scanf(%d,&f); c=(5*(f-32)/9; printf(%d

13、(F)=%.2f(C)n,f,c);3.13#include #define PI (3.1415926)int main(int argc, char* argv) double r, h; double s, v; printf(Please input the r and h.); scanf(%lf,%lf, &r, &h); s = 2 * PI * r * h + 2 * PI * r * r; v = PI * r * r * h; printf(s is %lf, v is %lf, s, v); return 0;3.14#include stdafx.hint main(i

14、nt argc, char* argv) char a4 = 编; printf(机内码：%x%xtn,a0&0xff,a1&0xff); printf(区位码：%xtn,a0&0xff8+a1&0xff-0x2020-0x8080); printf(国际码：%xtn,a0&0xff8+a1&0xff-0x8080); return 0;第四章习题4.1#include void main(void)float a,b,c;printf(Please enter the score of A:n);scanf(%f,&a);printf(Please enter the score of B:

15、n);scanf(%f,&b);printf(Please enter the score of C:n);scanf(%f,&c);if(a-b)*(a-c)0) printf(A gets the score %.1f,a);if(b-a)*(b-c)0) printf(B gets the score %.1f,b);if(c-a)*(c-b)0) printf(C gets the score %.1f,c);4.3#include int mdays(int y,int m) if (m=2) return (y%4=0 & (y%100=0 | y%400=0)?29:28; el

16、se if (m=4 | m=6 | m=9 | m=11) return 30; else return 31;main() int y,m,d,days; printf(Enter year:); scanf(%d,&y); printf(Enter month:); scanf(%d,&m); printf(Enter day:); scanf(%d,&d); days=d; while(m1)days+=mdays(y,m-1);m-; printf(%dn,days);4.4 if方法：#include stdafx.h#include int main(int argc, char

17、* argv) float x = 0; printf(input the salaryn); scanf(%f,&x); if(x0 & x1000) printf(0n); else if(x2000) printf(%fn,x*0.05); else if(x3000) printf(%fn,x*0.1); else if(x4000) printf(%fn,x*0.15); else if(x5000) printf(%fn,x*0.2); else printf(%fn,x*0.25); return 0;Case方法：#include stdafx.h#include int ma

18、in(int argc, char* argv) float x ; printf(input the salaryn); scanf(%f,&x); int xCase = 0; xCase = (int)(x/1000.0); switch(xCase) case 0: printf(0n); break; case 1: printf(%fn,x*0.05); break; case 2: printf(%fn,x*0.1); break; case 3: printf(%fn,x*0.15); break; case 4: printf(%fn,x*0.2); break; defau

19、lt: printf(%fn,x*0.25); return 0;4.7#include stdafx.h#include int main(int argc, char* argv) char *sa; char c; int i = 0,j = 0,k = 0; do c= getchar_r(); sai+ = c; while(c != r); for(i=0;sai+1;i+) for(j = i+1;saj;j+) if( sai=saj & saj = ) for(k=j;sak;k+) sak = sak+1; j-; for(k=0;sak;k+) printf(%2c,sa

20、k); return 0;4.10#include #define EPS 1e-5void main() int s=1; float n=1.0,t=1.0,pi=0; while(1.0/n=EPS) pi=pi+t; n=n+2; s=s*(-1); t=s/n; pi=pi*4; printf(pi=%10.6fn,pi);4.11#includeint main() int a,b,num1,num2,temp; printf(Input a & b:); scanf(%d%d,&num1,&num2); if(num1num2) temp=num1; num1=num2; num

21、2=temp; a=num1; b=num2; while(b!=0) temp=a%b; a=b; b=temp; printf(The GCD of %d and %d is: %dn,num1,num2,a); printf(The LCM of them is: %dn,num1*num2/a);4.13#include stdafx.h#include int Primes(int x);/判断素数函数int main(int argc, char* argv) int i,j; int num; for(num = 4;num=100;num+) if(num%2 = 0) for

22、(i=1;inum;i+) for(j=1;jnum;j+) if(num = i+j) if(Primes(i) & Primes(j) printf(%d=%d+%dn,num,i,j); return 0;int Primes(int x) int i ; int n = 0; for(i = 1;i=x;i+) if(x%i=0) n+; if(n=2) return 1; else return 0;4.17#includevoid main(void) int c,i; for(i=1,c=32;c=126;+i,+c) printf(%3d%-5c,c,c); if(!(i%8)

23、 printf(n); 4.18#include stdafx.h#include int main(int argc, char* argv) int x; int i,n,sum; printf(input 10 numbersn); for(i = 0,n = 0,sum = 0;i0) sum+=x; n+; if(n) printf(numbers = %d,average = %fn,n,1.0*sum/n); return 0;第五章习题5.5Extern和static存储类型的区别：Static型外部变量和extern型外部变量的唯一区别是作用域的限制。静态外部变量只能作用于定

24、义它的文件，其他文件中的函数不能使用。Extern型外部变量的作用域可以扩大到整个程序的所有文件。Static和auto存储类型的区别：静态局部变量和自动变量有根本性的区别。由于静态局部变量在程序执行期间不会消失，因此，它的值有连续性。当退出块时，它的值能保存下来，以便再次进入块时使用，而自动变量的值在退出块时都丢失了。如果定义时静态局部变量有显示初始化，只在第一次进入时执行一次赋初值操作，而自动变量每次进入时都要执行赋初值操作。5.6不能。在C语言中，参数的传递方式是“值传递”，即把实参的值拷贝到参数的存储区中。因此，swap()函数交换的只是实参的本地拷贝，代表swap()实参的变量并没有

25、被改变。5.7 6，125.10#include double sum_fac(int n) double s=0; int i; double fac=1.0; for(i=1;i=n;i+) fac*=1.0/i; s+=fac; return s;void main(void) int n; printf(Please enter the integer n:); scanf(%d,&n); printf(the sum is %lfn,sum_fac(n);5.17#include void reverse() char ch= getchar_r(); if(ch!=n) rever

26、se(); putchar(ch); int main() reverse(); printf(n); return 0;第六章习题6.1(1)正确(2)错误，不需要加“；”(3)错误，“ident”与“（”之间不能有空格(4)错误，宏名不能是关键字“void”(5)错误，将x*y改成(x）*(y)6.4将会导致变量blue的重复定义6.5(1)define NO 0(2)#include “common.h”(3)#line 3000(4)#undef TRUE #define TRUE 1(5)#if TRUE !=0#define FALSE 0#else #define FALSE 1

27、#endif(6)#ifndef SIZE assert(0);#elseassert(SIZE1);#endif(7)#define SQUARE_VOLUME(x) (x)*(x)*(x)6.10#include #define pi 3.1415926#define BALL_VOLUME(r) (4/3)*pi*(r)*(r)*(r)int main()int r;float v11;for(r=1;r11;r+)vr=float(BALL_VOLUME(r);printf(r=%2d v=%.2fn,r,vr);return 0;第七章习题7.9#include#include#de

28、fine g 10void main()char *buffer;int gdriver=DETECT,gmode,i,size;initgraph(&gdriver,&gmode,c:tcbgi);setbkcolor(BLUE);setcolor(RED);setlinestyle(0,0,1);setfillstyle(1,5);circle(200,250,RED);size=imagesize(200,250,200,300);buffer=malloc(size);getimage_r(200,250,200,300,buffer);for(i=0;i=10;i+)putimage

29、(200,250+g*i*i/2,buffer,COPY_PUT);getch_r();closegraph();7.11#include#define RAND_MAX 32767#define RAND 100int RandomInteger(int low,int high) int k; double d; d=(double)rand()/(double)RAND_MAX+1); k=(int)(d*(high-low+1); return(low+k);void main()long i;int n=0;int szWordRAND;char a=heads;char b=tai

30、ls;srand(time(0);for(i=0;iRAND;i+) szWordi=RandomInteger(0,1); if(szWordi=1) printf(n%s,a); n+; else printf(n%s,b); n=0; if(n=3) printf(nIt took %ld flips to get heads 3 consecutives times,i+1); break; 7.16char *MonthName(int month);int MonthDays(int month,int year);int FirstDayOfMonth(int month,int

31、 year);int IsLeapYear(int year);enum weakSUNDAY ,MONDAY,TUESDAY,WEDNESDAY,THURSDAY,FRIDAY,SATURDAY;#include “caltools.h”char *MonthName(int month)Switch(month)Case 1: return(“January”);Case 2: return(“February”);Case 3: return(“March”);Case 4: return(“April”);Case 5: return(“May”);Case 6: return(“Ju

32、ne”);Case 7: return(“July”);Case 8: return(“August”);Case 9: return(“September”);Case 10: return(“October”);Case 11: return(“November”);Case 12: return(“December”);Default : printf(“input error!n”);int MonthDays(int month,int year)Case 1:Case 3:Case 5:Case 7:Case 8:Case 10:Case 12: return(31);Case 4

33、:Case 6:Case 9:Case 11: return(30);Case 1: return (IsLeapYear(year)?29:28);int FirstDayOfMonth(int month,int year)Int i;Long total=0;If(year=2000)For(i=1;i2000)For (i=2000;iyear;i+)Total+=(IsLeapYear(i)?366:365);For(i=1;imonth;i-)Total+=( MonthDays(i, 1999);Return(13-total%7)%7);ElseFor (i=1999;imon

34、th;i-)Total+=( MonthDays(i, year);Return(13-total%7)%7);int IsLeapYear(int year)Return ( !(year%4)&(year%400) | !(year%400) );#include #include “caltools.h”Void main()Int month,year;Printf(“please input the month and year:”);Scanf(“%d%d” ,&month,&year);Printf(“the name of the month is %sn”, MonthNam

35、e( month);Printf(“there are %d days in this month.n”, MonthDays( month,int year);Printf(“the first day of the month in this year is %d”, FirstDayOfMonth( month,year);第八章习题8.4#include stdafx.h#include malloc.h#define N 65535void DelSpace(char sa);int main(int argc, char* argv) char saN; char c; int i

36、 =0; do c = getchar_r(); if(c = 13) sai+ = n; else sai+ = c; while(c!=); DelSpace(sa); int j = 0; while(1) if(saj = ) break; printf(%c,saj+); printf(/n); return 0;void DelSpace(char sa) char *t1 = (char*)malloc(sizeof(sa); char *t2 = (char*)malloc(sizeof(sa); t1 = t2 = sa; while(*t1) *t2+ = *t1+; if

37、(*(t2-1)= & *t1= ) t2-; 还有一个方法：void DelSpace(char sa,int n) char* tmpbuf = (char*)malloc(sizeof(sa)+1); int p1 = 0, p2 = 0; bool bSpace = false; while(p1 n) if(bSpace & sap1 = ) p1+; else if(sap1 = ) bSpace = true; else bSpace = false; tmpbufp2+ = sap1+; strcpy(sa, tmpbuf); free( tmpbuf);8.7#include

38、 stdafx.h#define NUM 100struct stuInfo char name10; int mark;stuNUM;void scoreSort(stuInfo stu,int n);int main(int argc, char* argv) int n; printf(input the number of students:n); scanf(%d,&n); printf(intput the name and scoren); for(int i = 0;in;i+) scanf(%s,&stui.name); scanf(%d,&stui.mark); score

39、Sort(stu,n); int j = 0; while(jn) printf(name:%s score:%dn,stuj.name,stuj.mark); j+; return 0;void scoreSort(stuInfo *stu,int n)/n为学生数 for(int i= 0;in;i+) for(int j =i;jn;j+) if(stui.markstuj.mark) stuInfo temp; temp = stui; stui = stuj; stuj = temp; 8.9/ c84.cpp : Defines the entry point for the co

40、nsole application./#include stdafx.h#define NUM 100struct stuInfo char name10; int mark;stuNUM;void scoreSort(stuInfo stu,int n);int find(stuInfo *stu,int n,int score);int main(int argc, char* argv) int n; printf(input the number of students:n); scanf(%d,&n); printf(intput the name and scoren); for(int i = 0;i=0) printf(name:%s score:%dn,stuifind.name,stuifind.mark); else printf(Not Find); return 0;void scoreSort(stuInfo *stu,int n)/n为学生数 for(in