Question

Following is the program that raised the mentioned doubt for me.

#include <stdio.h>
int main() { 
    int g = 300000*300000/300000;
    printf("%d",g);
    return 0;
}

When the * is evaluated the result would be 90000000000. Then is divided by 300000.
I expected the first expression result to be stored somewhere then divided by 300000. So output would be 300000.

But it is giving me -647.
Does this mean it is evaluated as :

g = 300000*300000;
g = g / 300000;

Answer 1

Regardless of where it’s stored, it’s still of type int. Assuming int is 32-bits on your machine, you’re getting integer overflow with 300000*300000.

300000*300000 -> 90000000000 -> -194313216  (integer overflow)
-194313216 / 300000 -> -647

Basically, temporaries (or intermediates) don’t magically allow you to get around overflow.

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*Note that signed integer overflow is technically undefined behavior. But in this case it happens to wrap-around the way you’d expect.