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copied!<p>By the way, this isn't computing factorial.</p> <p>If you're really trying to find out where the time goes, you could try <a href="https://stackoverflow.com/questions/375913/what-can-i-use-to-profile-c-code-in-linux/378024#378024">stackshots</a>. I put an infinite loop around your code and took 10 of them. Here's the code:</p> <pre><code> 6: void forloop(void){ 7: int fac=1; 8: int count=5; 9: int i,k; 10: 11: for (i = 1; i <= count; i++){ 12: for(k=1;k<=count;k++){ 13: fac = fac * i; 14: } 15: } 16: } 17: 18: int main(int argc, char* argv[]) 19: { 20: int i; 21: for (;;){ 22: forloop(); 23: } 24: return 0; 25: } </code></pre> <p>And here are the stackshots, re-ordered with the most frequent at the top:</p> <pre><code>forloop() line 12 main() line 23 forloop() line 12 + 21 bytes main() line 23 forloop() line 12 + 21 bytes main() line 23 forloop() line 12 + 9 bytes main() line 23 forloop() line 13 + 7 bytes main() line 23 forloop() line 13 + 3 bytes main() line 23 forloop() line 6 + 22 bytes main() line 23 forloop() line 14 main() line 23 forloop() line 7 main() line 23 forloop() line 11 + 9 bytes main() line 23 </code></pre> <p>What does this tell you? It says that line 12 consumes about 40% of the time, and line 13 consumes about 20% of the time. It also tells you that line 23 consumes nearly 100% of the time.</p> <p>That means unrolling the loop at line 12 might potentially give you a speedup factor of 100/(100-40) = 100/60 = 1.67x approximately. Of course there are other ways to speed up this code as well, such as by eliminating the inner loop, if you're really trying to compute factorial.</p> <p>I'm just pointing this out because it's a bone-simple way to do profiling.</p>

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