The compiler (in C) is allowed to assume that infinite loops eventually terminate. This can lead to these kinds of loops not actually running forever when built with an optimizing compiler.

ISO/IEC 9899:2017 §6.8.5 “Iteration statements”, paragraph 6:

“An iteration statement may be assumed by the implementation to terminate if its controlling expression is not a constant expression, and none of the following operations are performed in its body, controlling expression or (in the case of a for statement) its expression-3: – input/output operations – accessing a volatile object – synchronization or atomic operations."

It can, for example, simply optimize it away, assuming non-productive infinite loops are stupid and not reflective of what the code will actually do.

[–] azertyfun@sh.itjust.works 1 points 1 week ago (1 children)

if its controlling expression is not a constant expression

Pretty big caveat. If I'm reading this right true definitely qualifies as a constant expression and the loop in the meme would therefore not be optimized away.

[–] henfredemars@lemdro.id 2 points 1 week ago

There's also this part of the standard that throws a wrench into this hypothesis:

§5.1.2.3/4: (Program execution, Observable behavior):

Accesses to volatile objects and calls to library I/O functions are observable behavior. The implementation may perform any transformation of a program, provided that the resulting program’s observable behavior is not changed.

So it seems that running forever isn't an observable property that must be preserved when code is transformed.

Still, I think compilers try to not surprise the developer too badly and would recognize a trivial loop most of the time.

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