how much does array size affect compiling speed?

[larry burns]

thanks again for the suggestion!!! Heres the current code. I think the problem may be solved! I was able to get 500 n_simulations in only 5 minutes, which is about 10 times the previous speed. I'll now include the rest of the program that I didn't include earlier and see if it still works

[VladimirF]

You still have a few

{
    int t = 1;
}

left in your code. I thought you said that you've fixed it...

[VladimirF]

Few more comments?
Your code has a LOT of levels of nestedness (is that a word?), so whatever you can do to reduce it – please do.
Your statement:

if( n_simulations != 0 )

is right after n_simulations is initialized with 1 – just remove it.

I should have stated earlier – never EVER modify loop counter inside the loop (other then for regular step). You are making it impossible to read your code.

for ( double f = 0; f < 1; f = f + .004 )
{
    if( (x_value) < f + .004 && (x_value) >= f )
    {
        x = -(n_size - (2 * n_size * f) - 1);
        f = 1;
    }
}

is the same as

for ( double f = 0; f < 1; f = f + .004 )
{
    if( (x_value) < f + .004 && (x_value) >= f )
    {
        x = -(n_size - (2 * n_size * f) - 1);
        break;
    }
}

without forcing a reader to follow the usage of variable f to see why it needs to be set to 1.

Now look at this loop:

for ( int j = 1; j < 500; j++ )
{
    for ( int e = 0; e < 250; e++ )
    {
        for ( int f = 0; f < 250; f++ )
        {
            if( island[e][f] == j )
            {
                IslandSize[j - 1]++;
            }
        }
    }
}

What do you need the outer loop for? You are running through you 250*250 matrix (it’s already 62,500 iterations) 500 times, without changing anything!
Isn’t it pretty much the same as:

for ( int e = 0; e < 250; e++ )
{
    for ( int f = 0; f < 250; f++ )
    {
        IslandSize[island[e][f] - 1]++;
    }
}

only 500 times slower?

[Paul McKenzie]

In addition, this loop is indeterminate as to how many times it will execute:

for ( double f = 0; f < 1; f = f + .004 )

Never, never, never use doubles as loop counters!

The reason is that doubles are not exact. When you add 0.004, there is round-off error involved, since 0.004 cannot be represented exactly as a binary floating point number. So when you add 0.004, the value of f is not exact.

Now, how many times will the loop run? You don't know due to the round-off error.

Always use integers for loop counters. The loop should be rewritten:

double f;
for (int fInt = 0; fInt < 1000; fInt += 4)
{
   f = (double)fInt / 1000.0;
   // now do your calculations using f
//...
}

Regards,

Paul McKenzie