Are there extensions to let optimizers assume const-ref parameters will stay const?

Question

Related to my former question: Are compilers not allowed to assume const-ref parameters will stay const?

My new question is: Is there compiler-specific, non-standard ext

Answer 1

If i is supposed to stay const for the whole function, and f() has no side effects, you could declare it with __attribute__((pure)):

int f(const int&) __attribute__((pure));

Note that it doesn't make sense for a pure function to return void, so I changed it to int.

While this does not affect how f() is compiled, it does affect the functions calling it (check it on godbolt):

#include 

int f(const int& i) __attribute__((pure));

int main() {
    int i = 40;
    f(i);
    if (i != 40) {
        std::cout << "not 40" << std::endl;
    }
}

Here __attribute__((pure)) tells the compiler that f() will not change i, so the compiler will not generate the call to std::cout << ....

Without __attribute__((pure)), even if f() is declared to take a const int& i parameter, the compiler has to assume that the value of i may change, and generate the if and the call to std::cout << ... .