x+a%3*(int)(x+y)%2/4 //先算(int)(x+y),x+y取整得7
=x+a%3*7%2/4 //算a%3=1
=x+1*7%2/4 //算1*7=7
=x+7%2/4 //算7%2=1
=x+1/4 //算1/4=0
=x+0
=2.5/int x=2.5,a=7,y=4.7;//請(qǐng)問x，與y是整數(shù)嗎？//下面是測(cè)試程序#include int main(void){float x=2.5,y=4.7; float h; int a=7; h=x+a%3*(int)(x+y)%2/4; printf("%f\n",h);return 0;}