Why is taking the address of a temporary illegal?

Question

I know that the code written below is illegal

void doSomething(std::string *s){}
int main()
{
     doSomething(&std::string(\"Hello World\"));
     retur         


        

Answer 1

Long answer:

[...] it can be concluded that the temporary was present in some memory location

By definition:

• "temporary" stands for: temporary object
• an object occupies a region of storage
• all objects have an address

So it doesn't take a very elaborate proof to show that a temporary has an address. This is by definition.

OTOH, you are not just fetching the address, you are using the builtin address-of operator. The specification of the builtin address-of operator says that you must have a lvalue:

• &std::string() is ill-formed because std::string() is a rvalue. At runtime, this evaluation of this expression creates a temporary object as a side-effect, and the expression yield a rvalue that refers to the object created.
• &(std::string() = "Hello World") is well-formed because std::string() = "Hello World" is a lvalue. By definition, a lvalue refers to an object. The object this lvalue refers to is the exact same temporary

Short answer:

This is the rule. It doesn't need the (incorrect, unsound) justifications some people are making up.