lvalue-to-rvalue

TL;DR Given the following code: int* ptr; *ptr = 0; does *ptr require an lvalue-to-rvalue conversion of ptr before applying indirection? The standard covers the topic of lvalue-to-rvalue in many places but does not seem to specify enough information to determine whether the * operator require such a conversion. Details The lvalue-to-rvalue conversion is covered in N3485 in section 4.1 Lvalue-to-rvalue conversion paragraph 1 and says ( emphasis mine going forward ): A glvalue (3.10) of a non-function, non-array type T can be converted to a prvalue.53 If T is an incomplete type, a program that

I see the term "lvalue-to-rvalue conversion" used in many places throughout the C++ standard. This kind of conversion is often done implicitly, as far as I can tell. One unexpected (to me) feature of the phrasing from the standard is that they decide to treat lvalue-to-rvalue as a conversion. What if they had said that a glvalue is always acceptable instead of a prvalue. Would that phrase actually have a different meaning? For example, we read that lvalues and xvalues are examples of glvalues. We don't read that lvalues and xvalues are convertible to glvalues. Is there a difference in meaning?

I have a hard time understanding how this code (an example from the C++14 draft standard [conv.lval] ) invokes undefined behavior for g(false) . Why does constexpr make the program valid? Also, what does it mean by "does not access y.n "? In both calls to g() we are returning the n data member so why does the last line say it doesn't access it? struct S { int n; }; auto f() { S x { 1 }; constexpr S y { 2 }; return [&](bool b) { return (b ? y : x).n; }; } auto g = f(); int m = g(false); // undefined behavior due to access of x.n outside its // lifetime int n = g(true); // OK, does not access y