lvalue means “left value” — it should be assignable. You cannot change the value of text since it is an array, not a pointer. Either declare it as char pointer (in this case it’s better to declare it as const char*): const char *text; if(number == 2) text = “awesome”; else text = “you fail”; … Read more
Is bar an rvalue or an lvalue? The question answers itself. Whatever has a name is an lvalue(1). So bar is an lvalue. Its type is “rvalue reference to string“, but it’s an lvalue of that type. If you want to treat it as an rvalue, you need to apply std::move() to it. If you … Read more
Both i and j are glvalues (see this value category reference for details). Then if you read this conditional operator reference we come to this point: 4) If E2 and E3 are glvalues of the same type and the same value category, then the result has the same type and value category So the result … Read more
There is not much of a sane choice other than offering two overloads or making your function a template, I would say. If you really, really need an (ugly) alternative, then I guess the only (insane) thing you can do is to have your function accept a const&, with a pre-condition saying that you cannot … Read more
Most of the work is already done for you by the stdlib, you just need a function wrapper: template <typename T> constexpr bool is_lvalue(T&&) { return std::is_lvalue_reference<T>{}; } in the case you pass a std::string lvalue then T will deduce to std::string& or const std::string&, for rvalues it will deduce to std::string Note that Yakk’s … Read more
This is a gcc bug. The problem is that gcc incorrectly treats the expression !(!b) as an lvalue instead of rvalue. You can confirm this here. As you’ll see in the above linked demo, the output gcc gives is lvalue instead of prvalue. The bug has been reported as: GCC treats rvalue as an lvalue
No, the return value of a function is an l-value if and only if it is a reference (C++03). (5.2.2 [expr.call] / 10) If the type returned were a basic type then this would be a compile error. (5.17 [expr.ass] / 1) The reason that this works is that you are allowed to call member … Read more
Compilers are written in stages. The first stage is called the lexer and turns characters into a symbolic structure. So “++” becomes something like an enum SYMBOL_PLUSPLUS. Later, the parser stage turns this into an abstract syntax tree, but it can’t change the symbols. You can affect the lexer by inserting spaces (which end symbols … Read more
Compilers are written in stages. The first stage is called the lexer and turns characters into a symbolic structure. So “++” becomes something like an enum SYMBOL_PLUSPLUS. Later, the parser stage turns this into an abstract syntax tree, but it can’t change the symbols. You can affect the lexer by inserting spaces (which end symbols … Read more
Expressions don’t have return types, they have a type and – as it’s known in the latest C++ standard – a value category. A conditional expression can be an lvalue or an rvalue. This is its value category. (This is somewhat of a simplification, in C++11 we have lvalues, xvalues and prvalues.) In very broad … Read more