#include <stdio.h>
main() {
int a = 20;
int b = 10;
int c = 15;
int d = 5;
int e;
e = (a + b) * c / d; // ( 30 * 15 ) / 5
printf("Value of (a + b) * c / d is : %d/n", e );
e = ((a + b) * c) / d; // (30 * 15 ) / 5
printf("Value of ((a + b) * c) / d is : %d/n" , e );
e = (a + b) * (c / d); // (30) * (15/5)
printf("Value of (a + b) * (c / d) is : %d/n", e );
e = a + (b * c) / d; // 20 + (150/5)
printf("Value of a + (b * c) / d is : %d/n" , e );
return 0;
}
原创文章,作者:ItWorker,如若转载,请注明出处:https://blog.ytso.com/265128.html