What is efficient to check: equal to or not equal to? - c++

Question

I was wondering, if we have if-else condition, then what is computationally more efficient to check: using the equal to operator or the not equal to operator? Is there any difference at all?
E.g., which one of the following is computationally efficient, both cases below will do same thing, but which one is better (if there's any difference)?
Case1:
if (a == x)
{
// execute Set1 of statements
}
else
{
// execute Set2 of statements
}
Case 2:
if (a != x)
{
// execute Set2 of statements
}
else
{
// execute Set1 of statements
}
Here assumptions are most of the time (say 90% of the cases) a will be equal to x. a and x both are of unsigned integer type.

Answer 1

Generally it shouldn't matter for performance which operator you use. However it is recommended for branching that the most likely outcome of the if-statement comes first.

Answer 2

Usually what you should consider is; what is the simplest and clearest way to write this code? IMHO, the first, positive is the simplest (not requiring a !)
In terms of performance there is no differences as the code is likely to compile to the same thing. (Certainly in the JIT for Java it should)
For Java, the JIT can optimise the code so the most common branch is preferred by the branch prediction.

Answer 3

In this simple case, it makes no difference. (assuming a and x are basic types) If they're class-types with overloaded operator == or operator != they might be different, but I wouldn't worry about it.
For subsequent loops:
if ( c1 ) { }
else if ( c2 ) { }
else ...
the most likely condition should be put first, to prevent useless evaluations of the others. (again, not applicable here since you only have one else).

Answer 4

GCC provides a way to inform the compiler about the likely outcome of an expression:
if (__builtin_expect(expression, 1))
…
This built-in evaluates to the value of expression, but it informs the compiler that the likely result is 1 (true for Booleans). To use this, you should write expression as clearly as possible (for humans), then set the second parameter to whichever value is most likely to be the result.

Answer 5

There is no difference.
The x86 CPU architecture has two opcodes for conditional jumps
JNE (jump if not equal)
JE (jump if equal)
Usually they both take the same number of CPU cycles.
And even when they wouldn't, you could expect the compiler to do such trivial optimizations for you. Write what's most readable and what makes your intention more clear instead of worrying about microseconds.

Answer 6

If you ever manage to write a piece of Java code that can be proven to be significantly more efficient one way than the other, you should publish your result and raise an issue against whatever implementation you observed the difference on.
More to the point, just asking this kind of question should be a sign of something amiss: it is an indication that you are focusing your attention and efforts on a wrong aspect of your code. Real-life application performance always suffers from inadequate architecture; never from concerns such as this.

Answer 7

Early optimization is the root of all evil
Even for branch prediction, I think you should not care too much about this, until it is really necessary.
Just as Peter said, use the simplest way.

Answer 8

Let the compiler/optimizer do its work.
It's a general rule of thumb (most nowadays) that the source code should express your intention in the most readable way. You are writing it to another human (and not to the computer), the one year later yourself or your team mate who will need to understand your code with the less effort.

Answer 9

It shouldn't make any difference performance wise but you consider what is easiest to read. Then when you are looking back on your code or if someone is looking at it, you want it to be easy to understand.

Answer 10

it has a little advantage (from point of readability) if the first condition is the one that is true in most cases.
Write the conditions that way that you can read them best. You will not benefit from speed by negating a condition

Answer 11

Most processors use an electrical gate for equality/inequality checks, this means all bits are checked at once. Therefore it should make no difference, but you want to truly optimise your code it is always better to benchmark things yourself and check the results.
If you are wondering whether it's worth it to optimise like that, imagine you would have this check multiple times for every pixel in your screen, or scenarios like that. Imho, it is alwasy worth it to optimise, even if it's only to teach yourself good habits ;)

Answer 12

Only the non-negative approach which you have used at the first seems to be the best .

Answer 13

The only way to know for sure is to code up both versions and measure their performance. If the difference is only a percent or so, use the version that more clearly conveys the intent.
It's very unlikely that you're going to see a significant difference between the two.

Answer 14

Performance difference between them is negligible. So, just think about readability of the code. For readability I prefer the one which has a more lines of code in the If statement.
if (a == x) {
// x lines of code
} else {
// y lines of code where y < x
}