difftime returning 0 when there is clearly a difference

Question

I have the following C99 program which measures performance of simple division operations relative to addition. However, the difftime function keeps returning 0 even though the program is clearly taking several seconds to process runAddition and runDivision with iterations set to 1 billion.

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

void runAddition(long long iterations)
{
    long long temp;
    for (long long i = 1; i <= iterations; i++)
    {
        temp = temp + i;
    }
}

void runDivision(long long iterations)
{
    long long temp;

    // Start at 1 to avoid division by 0!
    for (long long i = 1; i <= iterations; i++)
    {
        temp = temp / i;
    }
}

int main()
{
    long long iterations = 1000000000;
    time_t startTime;

    printf("How many iterations would you like to run of each operation? ");
    scanf("%d", &iterations);

    printf("Running %d additions...\n", iterations);
    startTime = time(NULL);
    runAddition(iterations);
    printf("%d additions took %f seconds\n", iterations, difftime(time(NULL), startTime));

    printf("Running %d divisions...\n", iterations);
    startTime = time(NULL);
    runDivision(iterations);
    printf("%d divisions took %f seconds\n", iterations, difftime(time(NULL), startTime));
}

Answer 1

It calculates the difference in seconds between time1 and time2. So maybe your time difference is less than 1 second?

Output your start and end time to verify.

Answer 2

time() returns a time_t, which has a resolution of one second.

The time required for runDivision() is less than one second; one billion operations on a multi-GHz core will take less than a second.

Answer 3

Make temp volatile so it does not get optimized away. The compiler is probably seeing it as a section/function with no side effects.

Answer 4

Your format string expects an int (%d), and a double (%f). Your arguments are long long and double. You should set the first format string as %lld.

When pushing arguments on the stack to call printf, you push a long long using 8 bytes, and a double using 8 bytes too. When the function printf reads the format string, it expects first an int on 4 bytes, and a double on 8 bytes. printf gets the int correctly as you are little-endian and the first four bytes of your long long are enough to represent the value. printf then gets the double for which it gets the last four bytes of the long long, followed by the first four bytes of the double. As the last four bytes of the long long are zeroes, what printf thinks is a double starts with four bytes with value zero, resulting in a very very tiny value for the double according to the binary representation of doubles.

Answer 5

Try using %lld in place of %d in the printf:

printf("%lld additions t
         ^^^

Works fine after this change.