Simplify (bitset & flag) == flag conditions

[iluuu1994]

No description provided.

[TimWolla]

Was this a manual change or did you use Coccinelle for this? It looks like surprisingly few changes.

[iluuu1994]

@TimWolla I just used a regex ( & ([a-zA-Z]\w*)\) == \1). I've never tried Coccinelle :) There are a few more I found, but I only changed the code paths I help maintain (i.e. Zend/, ext/standard/, ext/opcache/, etc.).

Also, some of these hits are legitimate, namely when the constant contains multiple bits.

[TimWolla]

I'll try with Coccinelle when I don't forget, but I assume that the following patch would work:

@@
expression e;
expression b;
@@

- (e & b) == b
+ e & b

Possibly with some extra constraints to exclude locations where it's important that the result is a boolean.

What's the motivation of this patch? Readability? I would expect the compiler to optimize both variants into the same assembly.

[iluuu1994]

Readability, and the fact that it looks like (bitset & flags) == flags) but isn't meaningful as there is only one bit to begin with, which may be confusing.