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

&std::string("Hello World")

The problem with this isn't that std::string("Hello World") yields a temporary object. The problem is that the expression std::string("Hello World") is an rvalue expression that refers to a temporary object.

You cannot take the address of an rvalue because not all rvalues have addresses (and not all rvalues are objects). Consider the following:

42

This is an integer literal, which is a primary expression and an rvalue. It is not an object, and it (likely) does not have an address. &42 is nonsensical.

Yes, an rvalue may refer to an object, as is the case in your first example. The problem is that not all rvalues refer to objects.