Why does the execution result show a big difference in release mode and debug mode?

Question

# include <stdio.h>
# include<time.h>
# include <limits.h>

int main() {
    clock_t start;
    long a = 0;
    long b = 0;

    start = clock();
    for (int i = 0; i < INT_MAX; i++) {
        for (int j = 0; j < INT_MAX; j++) {
            for (int k = 0; k < INT_MAX; k++) {
                for (int q = 0; q < INT_MAX; q++) {
                    b = 1;
                }
            }
        }       
    }
    printf("%.5f\n", ((float)(clock() - start) / CLOCKS_PER_SEC));

    start = clock();
    for (int i = 0; i < INT_MAX; i++) {
        for (int j = 0; j < INT_MAX; j++) {
            for (int k = 0; k < INT_MAX; k++) {
                a = 0;
                for (int q = 0; q < INT_MAX; q++) {
                    a += 1;
                }
            }
        }
    }
    printf("%.5f\n",((float)(clock()-start)/CLOCKS_PER_SEC));   
}

When I run in release mode, this shows the result immediately. But when I run in Debug mode, it's not over.

I know that release mode is fast, but how can it be so fast?

An optimizing compiler will... well... optimize. In debug mode it won't go the same lengths. Why is it a shock? — StoryTeller - Unslander Monica, Nov 01 '18 at 06:23
@Sandeep read about it here: https://stackoverflow.com/questions/933739/what-is-the-difference-between-release-and-debug-modes-in-visual-studio — CIsForCookies, Nov 01 '18 at 06:44
@CIsForCookies I was trying to ask what the questioner meant by debug and release mode as there was no mention of VS, but thanks for the link. — Sandy, Nov 01 '18 at 07:36

score 6 · Answer 1 · answered Nov 01 '18 at 06:38

When optimizing the compiler may see that this code:

for (int i = 0; i < INT_MAX; i++) {
    for (int j = 0; j < INT_MAX; j++) {
        for (int k = 0; k < INT_MAX; k++) {
            for (int q = 0; q < INT_MAX; q++) {
                b = 1;
            }
        }
    }       
}

can simply be replaced by

b = 1;

Likewise the second loop block can be optimized away.

Further since a and b isn't used, the loops can be completely removed.

So your whole program may be optimized into something representing:

int main() {
    clock_t start;
    start = clock();
    printf("%.5f\n", ((float)(clock() - start) / CLOCKS_PER_SEC));
    start = clock();
    printf("%.5f\n",((float)(clock()-start)/CLOCKS_PER_SEC));   
}

GermanNerd · Answer 2 · 2018-11-01T11:57:02.160

To expand on this answer, even if you do use a and b, like in a printf statement after the nested loops, the loops can still be optimized away. Why?

Your loop limit is a constant which is known at compile time. So the compiler will actually do the calculation of what a will be in the end at compile time.

Try reading in a loop limit value from the console, and you will see that the second loop now will not be optimized away anymore.

But do not use a huge number like MAX_INT, because that will take very, very long - even on a fast machine. That is why you observe this:

But when I run in Debug mode, it's not over.

The program seems to run forever, right?

A number around 1000 will give you reasonable, measurable times.

Thank you for pointing that out. I changed my answer acccordingly. — GermanNerd, Nov 01 '18 at 11:57

Why does the execution result show a big difference in release mode and debug mode?

2 Answers2