函数练习参考代码教案.doc

试验九 函数 一、基础能力落实： 设计函数min(x,y),返回两个double数值中较小的数值。 #include stdio.h 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 stdio.h 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 = %5.2f\n”,num1,num2,add(num1,num2)); printf(“%5.2f + %5.2f = %5.2f\n”,num1,num2,subtract(num1,num2)); printf(“%5.2f + %5.2f = %5.2f\n”,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 stdio.h 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 stdio.h 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(“%.2f\n