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Home/ Questions/Q 8983553
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Editorial Team
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Editorial Team
Asked: June 15, 20262026-06-15T20:52:07+00:00 2026-06-15T20:52:07+00:00

I was wondering, if we have if-else condition, then what is computationally more efficient

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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.

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  1. Editorial Team
    Editorial Team
    2026-06-15T20:52:08+00:00Added an answer on June 15, 2026 at 8:52 pm

    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.

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