name-lookup

问题 It's not allowed to put a namespace and a class with the same name into one declarative region, i.e. namespace A {} class A{}; is ill-formed (see §3.3.1/4). However, one can introduce the name of either one via a using-directive: namespace N { namespace A {int i;} } struct A {static int i;}; using namespace N; int i = A::i; // The global struct, or namespace N::A? Is this code ill-formed? VC++ thinks so, as well as Clang: main.cpp:7:9: error: reference to 'A' is ambiguous int i = A::i; ^ main

问题 class messageA { }; class messageB { }; template<class T> class queue { public: virtual ~queue() {} void submit(T& x) {} }; class A : public queue<messageA>, public queue<messageB> { }; int main() { A aa; aa.submit(messageA()); aa.submit(messageB()); } My first thought is, the above code should be fine, as class A will contains 2 overloaded submit functions, which will accept messageA and messageB object. However, the compiler gives me the following error : May I know why there is an

问题 A point from the ISO C++ Draft n3290 : 3.4.0 2nd point A name “looked up in the context of an expression” is looked up as an unqualified name in the scope where the expression is found. Would someone please explain this statement with an example? 回答1: It says that the scope which contains the expression will be searched for the name. i.e. namespace foo { struct bar { void foobar() { do_something(); } }; } if you have this code the name do_something will be searched in the scope of foobar ,

问题 Consider this code: namespace A { int i = 24; } namespace B { using namespace A; int i = 11; int k = i; // finds B::i, no ambiguity } And basic.lookup.unqual.2: §6.4.1 Unqualified name lookup [basic.lookup.unqual] The declarations from the namespace nominated by a using-directive become visible in a namespace enclosing the using-directive; see [namespace.udir]. For the purpose of the unqualified name lookup rules described in [basic.lookup.unqual], the declarations from the namespace

问题 It recently came to my attention that member functions completely shadow free functions with the same name when inside the class. And by completely I mean that every free function with the same name is not considered for overload resolution at all. I can understand why it's done with something like this: void f(); struct S { void f(); void g() { f(); // calls S::f instead of ::f } }; where the functions have identical signatures, its only natural as variable scoping works the same way. But

问题 I'm writing some type traits to see if a free function exists with a specific set of parameters. The functions have a signature that looks something like this: template <class T> void func( SomeClass &, SomeType const & ); I know ahead of time the values for T , SomeClass , and SomeType . I want the trait to return true if this function exists with exactly these parameters, not using any implicit conversion. I can easily write some code to detect whether this function exists by using SFINAE

问题 Let us define f , as a friend function of S , inside the declaration of S : struct S { friend void f() {} }; I cannot find a way to call f . Is it true, then, that such an inline friend function can only be called with argument-dependant lookup? struct S { friend void f() {} friend void g(S const&) {} } const s; int main() { // f(); // error: 'f' was not declared in this scope // S::f(); // error: 'f' is not a member of 'S' g(s); // S::g(s); // error: 'g' is not a member of 'S' } Bonus: what

问题 I've been struggling with a compilation issue, and have been able to shrink the problem down to a small code segment. To set the stage, I'm trying to do CRTP, where the base method calls another in the derived class. The complication is, I want to use trailing return types to get the type of forwarding directly to the the Derived class's method. This always fails to compile unless I forward to the call operator in the derived class. This compiles: #include <utility> struct Incomplete;

问题 Suppose I define x as symbol function foo (defn foo [x] x) (def x foo) Can the name "foo" be discovered if only given x? Is there a way within foo to look up the name of the function x - "foo" in this case? (foo x) Is there or is it possible to create a function such as: (get-fn-name x) foo 回答1: A similar question was asked recently on this site; see here When you do (def x foo) , you are defining x to be "the value at foo ", not " foo itself". Once foo has resolved to its value, that value

问题 #include <stdio.h> int T; int main() { struct T { double x; }; printf("%zu", sizeof(T)); return 0; } If I run this code in C, the result is 4 , while in C++ it is 8 . Can someone explain why the difference? 回答1: Short answer: Because they aren't the same identifier, in fact. In C, structure names and variable names fall into different namespaces, so in C, sizeof(T) == sizeof(int) // global variable T sizeof(struct T) == sizeof(struct T) // not the same namespace In C++, however, structure