CODEWITHSUNDEEP
cperformance
curious
curious

Reputation: 1554

Why this huge difference in execution time happens for this tiny change into c code?

The below code in unix takes ~9s reported by time command.

int main()
{
    double u = 0;
    double v = 0;
    double w = 0;
    int i;
    for (i = 0;i < 1000000000;++i) {
        v *= w;
        u += v;
    }
    printf("%lf\n",u);
}

I don't understand why the execution times almost double when i change v *= w;withv *= u;

Upvotes: 2

Views: 116

Answers (3)

user1944441
user1944441

Reputation:

Compiler optimizes v*= w; to v = 0; and the probably u += v to u = 0; So those operations never happen.

Here is the test i did. Every version was done 10 times and averaged.

for (i = 0;i < 1000000000;++i) {
    v *= w;
    u += v;
}

4.0373 seconds



for (i = 0;i < 1000000000;++i) {
    v *= u;
    u += v;
}

7.3733 seconds



for (i = 0;i < 1000000000;++i) {
    v *= 0;
    u += 0;
}

4.0149 seconds

Upvotes: 1

Ganesh
Ganesh

Reputation: 5980

When you change v *= w to v *= u then there is an inter-dependency between the 2 statements. Hence, the first statement has to be completed before executing u += v which could be the reason for the increased performance as the compiler can't parallelize the execution.

Upvotes: 5

Montycarlo
Montycarlo

Reputation: 735

Probably because the compiler sees that w is never modified, and so can be compiled into a constant whereas the variable u is modified and so must have its own memory.

Upvotes: 2

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