Informative "if" statement in "for" loop

Question

Normally when I have a big for loop I put messages to inform me in which part of the process my program is, for example:

for(i = 0; i < large_n; i++) {
    if( i % (large_n)/1000 == 0) {
       printf("We are at %ld \n", i);
    }
    // Do some other stuff
}

I was wondering if this hurts too much the performance (a priori) and if it is the case if there is a smarter alternative.Thanks in advance.

Answer 1

Maybe you can split the large loop in order to check the condition sometimes only, but I don't know if this will really save time, that depends more on your "other stuff".

int T = ...; // times to check the condition, make sure large_n % T == 0
for(int t = 0; t < T; ++t)
{
  for(int i = large_n/T * t; i < large_n/T * (t+1); ++i)
  {
    // other stuff
  }
  printf("We are at %ld \n", large_n/T * (t+1));
}

Answer 2

Regardless of what is in your loop, I wouldn't be leaving statements like printf in unless it's essential to the application/user, nor would I use what are effectively redundant if statements, for the same reason.

Both of these are examples of trace level debugging. They're totally valid and in some cases very useful, but generally not ultimately so in the end application. In this respect, a usual thing to do is to only include them in the build when you actually want to use the information they provide. In this case, you might do something like this:

#define DEBUG

for(i = 0; i < large_n; i++) 
{
    #ifdef DEBUG
        if( i % (large_n)/1000 == 0) 
        {
            printf("We are at %ld \n", i);
        }
    #endif
}

Regarding the performance cost of including these debug outputs all the time, it will totally depend on the system you're running, the efficiency of whatever "printing" statement you're using to output the data, the check/s you're performing and, of course, how often you're trying to perform output.

Answer 3

Your mod test probably doesn't hurt performance but if you want a very quick test and you're prepared for multiples of two then consider a mathematical and test:

if ( ( i & 0xFF ) == 0 ) {
    /* this gets printed every 256 iterations */
    ...
}

or

if ( ( i & 0xFFFF ) == 0 ) {
    /* this gets printed every 65536 iterations */
    ...
}