The School of Hard Knocks is within each of us. Its experience, and millions of hours testing various algorithms to find the most efficient.

Depends on how much sleep you need.

You can always stick a finger in a powersocket and hope your brain gets somehow rearranged (improved), but my instinct is telling me that the change of succes will be greater if you just practice , practice and practice ;)

Cleaner?

for(int sort = 1; sort == 1;) {
        
        sort = 0;
        
        for(int i = 0; i < n - 1; i++) {
        
            if(a[i] < a[i+1]) {
            
                int temp = a[i];
                a[i] = a[i+1];
                a[i+1] = temp;
                
                sort = 1;
            }
        }
    }

a for loop works like this:

for ( init-expression ; cond-expression ; loop-expression )
{ 
    statement 
}

Your example only uses the init-expression amd cond-expression. So if 'sort' isn't changed in the statement-block this is an endless loop (you set 'sort' to 1, then loop while it's 1). This is normally written as: while(1) or for(;;)

Sure you can. DigitalPackrat sets sort to 0 in the loop, so it won't be endless. I was just trying to give you an insite on for-loops .
I prefer another way to write this loop:

int sorted = 0;
while (!sorted)
{
}

It reads a bit easier.

In niek_e's example, the init-expression, cond-expression, and loop-expression are all optional as long as you include the separating semicolons. That's why an infinite "forEVER" loop looks like this:

for (;;) {
  ...
}

As long as the loop does what you want, the syntax is very flexible. In Ed's experience, the usual culprits are a very long initialization that makes the loop line too long and a more complicated update:

// So long it makes more sense outside the loop
std::vector<std::pair<std::string, double> >::size_type index = 0;

for (; index != v.size(); ++index) {
  ...
}

// As a part of the loop (too long)
for (std::vector<std::pair<std::string, double> >::size_type index = 0; index != v.size(); ++index) {
  ...
}

// n isn't incremented with every loop
for (n = 0; n < 10 && std::getline(std::cin, input); ) {
  if (input[0] == '*')
    ++n;
}

Cool, huh? :)

Reiterating the point, niek_e made, its possible. All of them are not necessary, in fact none are.
Also, I did not want to use sort variable outside the loop, so I used it inside as, if there are more bubble sorts and I use the sort variable again and don't initialize it, well, that would give me some headache. Won't it?

Yup.