It seems to me that lvalue stands for "legal value", just a valid
expression in that context.
LS> A.Kazantsev wrote:
>> Agree. And what does the {lval} in the error message exactly mean?
LS> It probably means that the expression whose type was shown was
LS> an lvalue.
LS> An lvalue is something that can be assigned to, i.e. a writeable value.
LS> It's a weird name, but I think it's called that because it's something
LS> that can appear on the lefthand side of an assignment operator.
LS> In C/C++, not only can variables be lvalues (as in "a = 1"), but
LS> more complex expressions can be lvalues too. For example, this
LS> code will print "a is 1":
LS> int a, b;
LS> b = 5;
LS> (a = b) = 1;
LS> std::cout << "a is " << a << std::endl;
LS> What's going on is that "(a = b)" forms an expression that can
LS> be assigned to (an lvalue), and when you assign to that expression,
LS> you change the value of the variable a.
LS> As for why the compiler prints the "{lval}" in the error message
LS> for passing arguments to a function, I can't say for sure. At
LS> first, it seems a little odd because function arguments in C are
LS> by value only, so it wouldn't matter if it's an lvalue. But in
LS> C++, you can call by reference, so maybe it could matter.
LS> For instance:
LS> void increment (int& n)
LS> {
LS> n++;
LS> }
LS> int myfunction ()
LS> {
LS> int a;
LS> increment (a = 1);
LS> std::cout << "a is " << a << std::endl;
LS> }
LS> I might be wrong (I don't know the C++ specification by heart), but
LS> I believe this is valid C++ code and it prints "a is 2".
LS> What's going on is this:
LS> (1) increment() needs to be called, so...
LS> (2) "a = 1" is evaluated, causing the side-effect of modifying "a"
LS> to be 1. the value of the "a = 1" expression is the value of
LS> "a", which is numerically 1 and also an lvalue.
LS> (3) The lvalue is passed by reference to increment(), which
LS> then uses it to modify "a".
LS> Of course, this is very very bad style and highly confusing code,
LS> but the point is that if it's legal like I think it is, then the
LS> compiler might be putting "{lval}" in the error message on purpose,
LS> because it might actually mean something in some cases. Hmm,
LS> in particular it might be helpful if you try to pass a value
LS> which isn't an lvalue to a function that (because of a reference
LS> type on its argument) requires one.
LS> - Logan
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