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() {    return 3; }  void foo(int* ptr) {  }  int main() {    foo(&bar()); } 

In C++ it is not allowed to take the address of a temporary (or, at least, of an object referred to by an rvalue expression?), and I thought that this was because temporaries are not guaranteed to even have storage.

But then, although diagnostics may be presented in any form the compiler chooses, I'd still have expected MSVS to error rather than warn in such a case.

So, are temporaries guaranteed to have storage? And if so, why is the above code disallowed in the first place?

Answer 1

Actually, in the original language design it was allowed to take the address of a temporary. As you have noticed correctly, there is no technical reason for not allowing this, and MSVC still allows it today through a non-standard language extension.

The reason why C++ made it illegal is that binding references to temporaries clashes with another C++ language feature that was inherited from C: Implicit type conversion. Consider:

void CalculateStuff(long& out_param) {     long result;     // [...] complicated calculations     out_param = result; }  int stuff; CalculateStuff(stuff);  //< this won't compile in ISO C++ 

CalculateStuff() is supposed to return its result via the output parameter. But what really happens is this: The function accepts a long& but is given an argument of type int. Through C's implicit type conversion, that int is now implicitly converted to a variable of type long, creating an unnamed temporary in the process. So instead of the variable stuff, the function really operates on an unnamed temporary, and all side-effects applied by that function will be lost once that temporary is destroyed. The value of the variable stuff never changes.

References were introduced to C++ to allow operator overloading, because from the caller's point of view, they are syntactically identical to by-value calls (as opposed to pointer calls, which require an explicit & on the caller's side). Unfortunately it is exactly that syntactical equivalence that leads to troubles when combined with C's implicit type conversion.

Since Stroustrup wanted to keep both features (references and C-compatibility), he introduced the rule we all know today: Unnamed temporaries only bind to const references. With that additional rule, the above sample no longer compiles. Since the problem only occurs when the function applies side-effects to a reference parameter, it is still safe to bind unnamed temporaries to const references, which is therefore still allowed.

This whole story is also described in Chapter 3.7 of Design and Evolution of C++:

The reason to allow references to be initialized by non-lvalues was to allow the distinction between call-by-value and call-by-reference to be a detail specified by the called function and of no interest to the caller. For const references, this is possible; for non-const references it is not. For Release 2.0 the definition of C++ was changed to reflect this.

I also vaguely remember reading in a paper who first discovered this behavior, but I can't remember right now. Maybe someone can help me out?

Answer 2

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

template<typename T> 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<typename T> 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.