c++ - Does the compiler optimize the function parameters passed by value?

Question

Lets say I have a function where the parameter is passed by value instead of const-reference. Further, lets assume that only the value is used inside the function i.e. the function doesn't try to modify it. In that case will the compiler will be able to figure out that it can pass the value by const-reference (for performance reasons) and generate the code accordingly? Is there any compiler which does that?

Answer 1

If you pass a variable instead of a temporary, the compiler is not allowed to optimize away the copy if the copy constructor of it does anything you would notice when running the program ("observable behavior": inputs/outputs, or changing volatile variables).

Apart from that, the compiler is free to do everything it wants (it only needs to resemble the observable behavior as-if it wouldn't have optimized at all).

Only when the argument is an rvalue (most temporary), the compiler is allowed to optimize the copy to the by-value parameter even if the copy constructor has observable side effects.