Why is it illegal to take the address of an rvalue temporary?

Question

According to \" How to get around the warning "rvalue used as lvalue"? \", Visual Studio will merely warn on code such as this:

int bar() {
   retu         


        

Answer 1

Certainly temporaries have storage. You could do something like this:

template
const T *get_temporary_address(const T &x) {
    return &x;
}

int bar() { return 42; }

int main() {
    std::cout << (const void *)get_temporary_address(bar()) << std::endl;
}

In C++11, you can do this with non-const rvalue references too:

template
T *get_temporary_address(T &&x) {
    return &x;
}

int bar() { return 42; }

int main() {
    std::cout << (const void *)get_temporary_address(bar()) << std::endl;
}

Note, of course, that dereferencing the pointer in question (outside of get_temporary_address itself) is a very bad idea; the temporary only lives to the end of the full expression, and so having a pointer to it escape the expression is almost always a recipe for disaster.

Further, note that no compiler is ever required to reject an invalid program. The C and C++ standards merely call for diagnostics (ie, an error or warning), upon which the compiler may reject the program, or it may compile a program, with undefined behavior at runtime. If you would like your compiler to strictly reject programs which produce diagnostics, configure it to convert warnings to errors.