C语言中的算数表达式设x=2.5,a=7,y=4.7 则x+a%3*(int)(x+y)%2/4的值为多少?
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![C语言中的算数表达式设x=2.5,a=7,y=4.7 则x+a%3*(int)(x+y)%2/4的值为多少?](/uploads/image/z/11811-3-1.jpg?t=C%E8%AF%AD%E8%A8%80%E4%B8%AD%E7%9A%84%E7%AE%97%E6%95%B0%E8%A1%A8%E8%BE%BE%E5%BC%8F%E8%AE%BEx%3D2.5%2Ca%3D7%2Cy%3D4.7+%E5%88%99x%2Ba%253%2A%28int%29%28x%2By%29%252%2F4%E7%9A%84%E5%80%BC%E4%B8%BA%E5%A4%9A%E5%B0%91%3F)
C语言中的算数表达式设x=2.5,a=7,y=4.7 则x+a%3*(int)(x+y)%2/4的值为多少?
C语言中的算数表达式
设x=2.5,a=7,y=4.7 则x+a%3*(int)(x+y)%2/4的值为多少?
C语言中的算数表达式设x=2.5,a=7,y=4.7 则x+a%3*(int)(x+y)%2/4的值为多少?
答案是2.500000.那个int(x+y)强制类型转换成整型,结果是7;前面a%3求余为1,即1*7%2/4;7与2求余是1,与4整除为0;则答案是x的值. 程序如下
#include
int main()
{
float x=2.5;
i nt a=7;
float sum;
float y=4.7;
sum = x+a%3*(int)(x+y)%2/4;
printf("%f\n", sum);
return 0;
}