CppCheck warning: expression depends on order of evaluation in x = x |= (1 << 3)

Question

The line of code in C is

x = x |= (1 << 3);

which gives an cppCheck Error: "Expression 'x=x|=1' depends on order of evaluation of side effects"

whereas the line

x |= (1 << 3);

is ok.

I thought

x = x |= (1 << 3);

whould be the same as

x = x = x | (1 << 3);

which is just

x = (x = (x | (1 << 3)));

where actually the outer assignment to x has no effect, meaning the outcome is the same as

x |= (1 << 3);

So what exactly is CppCheck complaining about here?

edit: think it is a duplicate of why j = j++ is or is not the same as j++ which is discussed in the question referred to above.

Answer 1

This quote from @Cornstalks' link on sequence points explains it very well.

Expressions ... which modify the same value twice are abominations which needn't be allowed (or in any case, needn't be well-defined, ie we don't have to figure out a way to say what they do, and compilers don't have to support them).

The C Standard simply does not mandate anything about these types of expressions, and therefore there is no particular order of evaluation that is guaranteed in all environments.

a rather quick&&simple explanation:

x = x |= 1 is pretty much equivalent to x = x += 1 in terms of side effects(modifications to x). x = x += 1 is equivalent to x = ++x in C. This expression is a well-known undefined expression .

Read more about it [here]