https://gcc.gnu.org/bugzilla/show_bug.cgi?id=98710
--- Comment #8 from Ivan Sorokin <vanyacpp at gmail dot com> --- > How often these show up, I have no idea. Perhaps I should have written this in the original message. The original expression "(x | c) & ~(y | c)" is obviously a reduced version of what happens in real code. The idea is the following: When we are working with bitsets "|" can be read as adding bits and "&~" as removing. Therefore the expression can be read as first adding "c" to "x" and then removing "y | c" from the result. So the simplification (x | c) & ~(y | c) -> x & ~(y | c) means there is no need to add "c" if later we remove something containing "c".
