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Home/ Questions/Q 799605
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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T23:09:21+00:00 2026-05-14T23:09:21+00:00

I have the following code that I’m reading through: if( (i%2) == 0 ){

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I have the following code that I’m reading through:

if( (i%2) == 0 ){ 
    *d = ((b & 0x0F) << 4); 
}
else{
    *d++ |= (b & 0x0F); 
};

I’m looking specifically at the else statement and wondering in what order this occurs? I don’t have a regular C compiler, so I can’t test this. When we are performing *d++ |= (b & 0x0F);, what order does this occur in?

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  1. Editorial Team
    Editorial Team
    2026-05-14T23:09:22+00:00Added an answer on May 14, 2026 at 11:09 pm

    The ++ is applied on the pointer d, not on the lvalue that is being assigned to, *d.

    If you really want to, you can think of it like this:

    1. The value of b is bitwise-AND:ed with the constant 0x0f
    2. The resulting bit pattern is bitwise-OR:ed into the value that d points at.
    3. The pointer d is incremented to point at the next value.
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