函数练习参考代码

试验九 函数一、基础能力落实：1) 设计函数min(x,y),返回两个double数值中较小的数值。#include double min(double a, double b);int main(void) double x, y; printf(Enter two numbers: ); scanf(%lf %lf, &x, &y); printf(The smaller number is %f.n, min(x,y); return 0;double min(double a, double b) return a b ? a : b;2) 编写一个程序，调函数实现功能：(1)求两个数之和；(2)求两个数之差；（3）求两个数之积。#include double add(double a, double b)return a+b;double subtract(double a, double b)return a-b;double multiple(double a,double b)return a*b;int main()double num1,num2;printf(“please input tow number:”);scanf(“%lf %lf”,&num1,&num2);printf(“%5.2f + %5.2f = %5.2fn”,num1,num2,add(num1,num2);printf(“%5.2f + %5.2f = %5.2fn”,num1,num2,subtract(num1,num2);printf(“%5.2f + %5.2f = %5.2fn”,num1,num2,multiple(num1,num2);3) 编写一个函数，函数的3个参数是一个字符和两个整数。字符参数是需要输出的字符，第一个整数说明了在每行中该字符输出的个数，而第二个整数指的是需要输出的行数，编写一个调用该函数的程序。#include void chLineRow(char ch, int c, int r);int main(void) char ch; int col, row; printf(Enter a character (# to quit): ); while ( (ch = getchar() != #) if (ch = n) continue; printf(Enter number of columns and number of rows: ); if (scanf(%d %d, &col, &row) != 2) break; chLineRow(ch, col, row); printf(nEnter next character (# to quit): ); printf(Bye!n); return 0;void chLineRow(char ch, int c, int r) int col, row; for (row = 0; row r ; row+) for (col = 0; col c; col+) putchar(ch); putchar(n); 4）以下函数的功能是用递归的方法计算x的n阶勒让德多项式的值。已有调用语句p（n，x）；编写函数实现功能。递归公式如下：#include float p(int x,int n)float t,t1,t2;if(n= =0) return 1;else if(n= =1) return x;else t1=(2*n-1)*x*p(x,(n-1);t2=(n-1)*p(x,(n-2);t=(t1-t2)/n;return t;int main()int x,n;printf(“input two int (x and n):”);scanf(“%d%d”,&x,&n);printf(“%.2fn”,p(x,n);return 0;5）编写一个函数taxis（）实现数组的排序，在函数中调用swap（）实现两个数的交换。打印出排序结果。#include void swap(int *a,int *b)int tmp;tmp=*a;*a=*b;*b=tmp;void taxis(int *a, int num)int i,j;for(i=0;inum-1;i+)for(j=i+1;j aj)swap(&ai,&aj);int main()int i,a=2,5,9,8,7,6,4,3,1;taxis(a, sizeof(a)/sizeof(int);for(i=0;isizeof(a)/sizeof(int);i+)printf(“%dt”,ai);printf(“n”);return 0;二、进阶能力落实：1) 编写一个函数，实现两个字符串的比较。#include int my_strcmp(char *p,char *q)int i=0;while(*(p+i) = *(q+i) if( *(p+i) = =0 ) return 0; i+; return(*(p+i) - *(q+i);int main()int m;char str120,str220;printf(“Input tow strings:”);gets(str1); gets(str2);printf(“result:%dn”,my_strmp(str1,str2); return 0;2) 编写一个函数is-within().它接受两个参数，一个是字符，另一个是字符串指针。其功能是如果字符在字符串中。就返回一个非0值（真）；如果字符不在字符串中，就返回0值（假）。在一个使用循环语句为这个函数提供输入的完整程序中进行测试。#include #define LEN 80int is_within(const char * str, char c);int main(void) char inputLEN; char ch; int found; printf(Enter a string: ); while (gets(input) & input0 != 0) printf(Enter a character: ); scanf(“n%c”, &ch); found = is_within(input, ch); if (found = = 0) printf(%c not found in string.n, ch); else printf(%c found in string %sn, ch, input); printf(Next string: ); puts(Done.n); return 0;int is_within(const char * str, char ch) while (*str != ch & *str != 0) str+; return *str; /* = 0 if 0is reached, non-zero otherwise */3) 有一个班，有3个学生，各四门课，计算总平均分数，以及地n个学生的成绩。（用函数average求平均成绩，用函数search找出并输出第i个学生的成绩）int main()int num;float score34=65,67,70,60,80,87,90,81,90,99,100,98;printf(“average=%5.2fn”,average(score0,12);while(1)printf(“Input a number(13):”);scanf(“%d”,&num);if(num 3) | (num 1)printf(“error!n”);else search(score,num);return 0;float average(float *p,int n)float *p_end;float sum=0,aver;p_end=p+n-1;for(;p=p_end;p+)sum+=*p;aver=sum/n;return aver;int search(float (*p)4,int n)int i;printf(“the score of NO.%d are:”,n);for(i=0;i4;i+)printf(“%2.0f”,*(*(p+n-1)+i);return 0;4) 编写并测试函数larger_of()，其功能是将两个double类型变量的数值替换成它们中的较大值，例如：larger_of(x,y)会把x,y中的较大数值重新赋给变量x和y.#inlcude void larger_of(double *p1, double *p2);int main(void) double x, y; printf(Enter two numbers (q to quit): ); while (scanf(%lf %lf, &x, &y) = 2) larger_of(&x, &y); printf(The modified values are %f and %f.n, x, y); printf(Next two values (q to quit): ); printf(Bye!n); return 0;void larger_of(double *p1, double *p2) double temp = *p1 *p2 ? *p1 : *p2; *p1= *p2 = temp;5) 编写一个函数，其参数为一个字符串，函数删除字符串中的空格，在一个可以循环读取的程序中进行测试，直到用户输入空行，对于任何输入字符串，函数都应该使用和并可以显示结果#include #define LEN 81int drop_space(char * s);int main(void) char origLEN; while (gets(orig) & orig0 != 0) drop_space(orig); puts(orig); puts(Bye!); return 0;int d