12、设变量x为float型且已经赋值,则以下语句中能够将x中的数值保留到小数点后面2位,并将第3位四舍五入的是__B__.A)x=x*100+0.5/100.0 B)x=(x*100+0.5)/100.0C)x=(int)(x*100+0.5)/100.0 D)x=(x/100+0.5)*100.0我要分析过程,谢谢大虾...
include <conio.h> include <windows.h> float fun ( float h ){ long num;h=h+(float)0.005;h=h*100;num=(long)h;h=(float)num;h=h/100;return h;} main(){ float a;system("cls");printf("Enter a: ");scanf("%f",&a);printf("The original data is: ");printf("%f ...
include "stdio.h"float fun(double h){return (int((h+0.005)*100))/100.0f;}int main(int argv,char *argc[]){//float a;double a;//clrscr();printf("Enter a:");//scanf("%f",&a);scanf("%lf",&a);printf("The original data is:%f\n",a);//printf("The result :...
答案 C (int)(x*100+0.5) 把float型数据(x*100+0.5)强转成int,这样就可以去掉小数点,+0.5就是为了四舍五入 例如x=4.256,则(4.256*100+0.5)=426.1 ,则(int)426.1/100.0=4.26 你选的B是错误的,没有强转成int,所以426.1/100=4.261,并没有达到题目要求 ...