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;
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.
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.
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
v *= u; since is 0 as well
v is constant zero, then that u never changes and remains zero - in the other case, it just has to see that w is constant zero), that could make a difference.
wis always zero. The compiler is probably smart enough not to run that loop at all in the first case sincevwill also always be zero, and thusutoo. (Might be harder to figure out with the interdependency between u and v.)