Question

The code I have looks like this (all uses of done shown):

bool done = false;
for(int i = 0; i < big; i++)
{
  ...
  for(int j = 0; j < wow; j++)
  {
    ...
    if(foo(i,j))
    {
       done = true;
       break;
    }
    ...
  }
  if(done) break;
  ...
}

will any compilers convert it to this:

for(int i = 0; i < big; i++)
{
  ...
  for(int j = 0; j < wow; j++)
  {
    ...
    if(foo(i,j))
      goto __done; // same as a labeled break if we had it
    ...
  }
  ...
}
__done:;

Note: While I’m mostly interested in if the if(done)break; gets bypassed and removed as dead code, I’m also interested in if it and done gets removed altogether.

Answer 1

Obviously this depends on the compiler. The best thing to do when you’re unsure is to view the compiler’s assembly output (all popular compilers have a switch for this). Even if you aren’t familiar with assembly, you can at least compare the debug version with the optimized version.

That being said, this is one of the few situations where goto is NOT a bad idea. Feel free to use it to break out of inner loops.

Edit

Just tried the following in VS2010 and it does indeed optimize the outer conditional:

bool done = false;
for(int i = 0; i < 10; i++)
{
    for(int j = 0; j < 10; j++)
    {
        if(i == 7 && j == 3)
        {
            done = true;
            break;
        }
    }
    if(done) break;
}
return 0;