I think the lvalue-to-rvalue conversion is more than just use an lvalue where an rvalue is required. It can create a copy of a class, and always yields a value, not an object.
I'm using n3485 for "C++11" and n1256 for "C99".
Objects and values
The most concise description is in C99/3.14:
object
region of data storage in the execution environment, the contents of which can represent
values
There's also a bit in C++11/[intro.object]/1
Some objects are polymorphic; the implementation generates information associated with
each such object that makes it possible to determine that object’s type during program execution. For other objects, the interpretation of the values found therein is determined by the type of the expressions used to access them.
So an object contains a value (can contain).
Value categories
Despite its name, value categories classify expressions, not values. lvalue-expressions even cannot be considered values.
The full taxonomy / categorization can be found in [basic.lval]; here's a StackOverflow discussion.
Here are the parts about objects:
- An lvalue ([...]) designates a function or an object. [...]
- An xvalue (an “eXpiring” value) also refers to an object [...]
- A glvalue (“generalized” lvalue) is an lvalue or an xvalue.
- An rvalue ([...]) is an xvalue, a temporary object or subobject thereof, or a value that is not associated with an object.
- A prvalue (“pure” rvalue) is an rvalue that is not an xvalue. [...]
Note the phrase "a value that is not associated with an object". Also note that as xvalue-expressions refer to objects, true values must always occur as prvalue-expressions.
The lvalue-to-rvalue conversion
As footnote 53 indicates, it should now be called "glvalue-to-prvalue conversion". First, here's the quote:
1 A glvalue of a non-function, non-array type T can be converted to a prvalue. If T is an incomplete type, a program that necessitates this conversion is ill-formed. If the object to which the glvalue refers is not an object of type T and is not an object of a type derived from T, or if the object is uninitialized, a program
that necessitates this conversion has undefined behavior. If T is a non-class type, the type of the prvalue is the cv-unqualified version of T. Otherwise, the type of the prvalue is T.
This first paragraph specifies the requirements and the resulting type of the conversion. It isn't yet concerned with the effects of the conversion (other than Undefined Behaviour).
2 When an lvalue-to-rvalue conversion occurs in an unevaluated operand or a subexpression thereof the value contained in the referenced object is not accessed. Otherwise, if the glvalue has a class type, the conversion copy-initializes a temporary of type T from the glvalue and the result of the conversion is a prvalue for the temporary. Otherwise, if the glvalue has (possibly cv-qualified) type std::nullptr_t, the
prvalue result is a null pointer constant. Otherwise, the value contained in the object indicated by the glvalue is the prvalue result.
I'd argue that you'll see the lvalue-to-rvalue conversion most often applied to non-class types. For example,
struct my_class { int m; };
my_class x{42};
my_class y{0};
x = y;
The expression x = y does not apply the lvalue-to-rvalue conversion to y (that would create a temporary my_class, by the way). The reason is that x = y is interpreted as x.operator=(y), which takes y per default by reference, not by value (for reference binding, see below; it cannot bind an rvalue, as that would be a temporary object different from y). However, the default definition of my_class::operator= does apply the lvalue-to-rvalue conversion to x.m.
Therefore, the most important part to me seems to be
Otherwise, the value contained in the object indicated by the glvalue is the prvalue result.
So typically, an lvalue-to-rvalue conversion will just read the value from an object. It isn't just a no-op conversion between value (expression) categories; it can even create a temporary by calling a copy constructor. And the lvalue-to-rvalue conversion always returns a prvalue value, not a (temporary) object.
Note that the lvalue-to-rvalue conversion is not the only conversion that converts an lvalue to a prvalue: There's also the array-to-pointer conversion and the function-to-pointer conversion.
values and expressions
Most expressions don't yield objects[[citation needed]]. However, an id-expression can be an identifier, which denotes an entity. An object is an entity, so there are expressions which yield objects:
int x;
x = 5;
The left hand side of the assignment-expression x = 5 also needs to be an expression. x here is an id-expression, because x is an identifier. The result of this id-expression is the object denoted by x.
Expressions apply implicit conversions: [expr]/9
Whenever a glvalue expression appears as an operand of an operator that expects a prvalue for that operand, the lvalue-to-rvalue, array-to-pointer, or function-to-pointer standard conversions are applied to convert the expression to a prvalue.
And /10 about usual arithmetic conversions as well as /3 about user-defined conversions.
I'd love now to quote an operator that "expects a prvalue for that operand", but cannot find any but casts. For example, [expr.dynamic.cast]/2 "If T is a pointer type, v [the operand] shall be a prvalue of a pointer to complete class type".
The usual arithmetic conversions required by many arithmetic operators do invoke an lvalue-to-rvalue conversion indirectly via the standard conversion used. All standard conversions but the three that convert from lvalues to rvalues expect prvalues.
The simple assignment however doesn't invoke the usual arithmetic conversions. It is defined in [expr.ass]/2 as:
In simple assignment (=), the value of the expression replaces that of the object referred to by the left operand.
So although it doesn't explicitly require a prvalue expression on the right hand side, it does require a value. It is not clear to me if this strictly requires the lvalue-to-rvalue conversion. There's an argument that accessing the value of an uninitialized variable should always invoke undefined behaviour (also see CWG 616), no matter if it's by assigning its value to an object or by adding its value to another value. But this undefined behaviour is only required for an lvalue-to-rvalue conversion (AFAIK), which then should be the only way to access the value stored in an object.
If this more conceptual view is valid, that we need the lvalue-to-rvalue conversion to access the value inside an object, then it'd be much easier to understand where it is (and needs to be) applied.
Initialization
As with simple assignment, there's a discussion whether or not the lvalue-to-rvalue conversion is required to initialize another object:
int x = 42; // initializer is a non-string literal -> prvalue
int y = x; // initializer is an object / lvalue
For fundamental types, [dcl.init]/17 last bullet point says:
Otherwise, the initial value of the object being initialized is the (possibly converted) value of the initializer expression. Standard conversions will be used, if necessary, to convert the initializer expression to the cv-unqualified version of the destination type; no user-defined conversions are considered. If the conversion cannot be done, the initialization is ill-formed.
However, it also mentioned the value of the initializer expression. Similar to the simple-assignment-expression, we can take this as an indirect invocation of the lvalue-to-rvalue conversion.
Reference binding
If we see lvalue-to-rvalue conversion as a way to access the value of an object (plus the creation of a temporary for class type operands), we understand that it's not applied generally for binding to a reference: A reference is an lvalue, it always refers to an object. So if we bound values to references, we'd need to create temporary objects holding those values. And this is indeed the case if the initializer-expression of a reference is a prvalue (which is a value or a temporary object):
int const& lr = 42; // create a temporary object, bind it to `r`
int&& rv = 42; // same
Binding a prvalue to an lvalue reference is prohibited, but prvalues of class types with conversion functions that yield lvalue references may be bound to lvalue references of the converted type.
The complete description of reference binding in [dcl.init.ref] is rather long and rather off-topic. I think the essence of it relating to this question is that references refer to objects, therefore no glvalue-to-prvalue (object-to-value) conversion.
int && y = x;isn't an expression itself, but a declaration. Therefore, lvalue-to-rvalue conversion isn't automatically applied to the "operand on the right side of=". – Postmarkint x = 42; int && y = +x;(the unary+invokes lvalue-to-rvalue conversion) -- this however doesn't bindytox, but to a temporary, as it had forint && y = 42;. – Postmark1, it would have worked. Other than a reference declaration, is there any place where we cannot replace a constant, with an lvalue, and have the code still work? In the above context, we cannot do that (we can usestd::move(), as you say, but I realize now that I am interested in the failure of implicit lvalue-to-rvalue conversion (I've amended the question)). – Choint && x = 1; ++x;doesn't actually change the value of1. Butint y = 42; int &&x = std::move(y); ++x;will change the value ofy. – Hatterasint x; int&& r = x;would bindrto a temporary/copy ofx. If it wouldn't bind a temporary butxdirectly, that would be dangerous, as rvalue references conventionally imply that you can steal the resources of the objects they refer to. – Postmark