printf中的C语言计算

Question

#include <stdio.h>

int main(){
int x;
x = (10+20)*(1.0/2);
printf("%d", x);
return 0;
}

the output is

15

and in another form:

#include <stdio.h>

int main(){
printf("%d", (10+20)*(1.0/2));
}

the output is

0

1.0/2 is 0.5 when calculate in integer variable x , but 1.0/2 is 0 when calculate in printf with %d (I think...)

I cannot understand this situation...

In addition, when I change %d to %f in the second code, the output is 15.000000

I'm using DEV C++ in Windows 8.

Answer 1

The type of (10+20)*(1.0/2) is double , but "%d" expects an int .

So, you need to cast to int :

printf("%d", (int) ((10+20)*(1.0/2)));

It works with "%f" because that expects a double .

Answer 2

as said before, your problem is casting issues. since 1.0/2 = 0.5 --> it is translated to 0, as a result of int casting. the right way to do it is either using a calculation function - that returns an integer, casting it to int (as proposed), or doing the calculations before the printf.