Question

Given:

int main() {
   int x = 0;
   int y = x; // <---
}

Could someone please tell me which clause of the standard (2003 preferred) mandates the conversion of the expression x from lvalue to rvalue in the initialisation of the object y?

(Or, if I’m mistaken and no such conversion takes place, then I’d like to learn that too!)

Answer 1

I find it easier (if maybe not 100% precise) to think of lvalue-s as real objects and rvalue-s as the value stored in the object. The expression x is an lvalue expression that refers to the object x defined in the first line, but when used as the right hand side of an assignment to a type that is not a user defined type the actual value is read, and that is where the conversion from lvalue to rvalue is performed: reading the contents of the object.

As to the specific clause in the standard that dictates that conversion… well, the closest that I can think is 4.1 [conv.lvalue]/2 (Lvalue to Rvalue conversion):

The value contained in the object indicated by the lvalue is the rvalue result.

The requirement that the right hand side of the assignment is an rvalue is either implicit or missing from 5.17 [expr.ass], but that is the case or else the following expression would be an error since the rhs is an rvalue and there is no rvalue-to-lvalue conversion:

int x = 5;

EDIT: For initialization, 8.5 [dcl.init]/14, last bullet (which refers to fundamental types) states (emphasis mine):

• Otherwise, the initial value of the object being initialized is the (possibly converted) value of the initializer expression. […]

That value there means that the lvalue expression in your example is read (i.e. converted to an rvalue). At any rate the previous paragraph that referred to assignment could be applied here: if initialization required an lvalue rather than an rvalue, the expression int i = 0; would be ill-formed.