任何函数都可以被删除的函数？ [英] Can any function be a deleted-function?

问题描述

工作草案明确调用 defaulted-functions 必须是特殊成员函数（例如，copy-constructor，default-constructor等，§8.4.2.1-1）。

The working draft explicitly calls out that defaulted-functions must be special member functions (eg copy-constructor, default-constructor, etc, (§8.4.2.1-1)). Which makes perfect sense.

但是，我没有在 deleted-functions （§8.4.3）中看到任何这样的限制。是这样吗？

However, I don't see any such restriction on deleted-functions(§8.4.3). Is that right?

或者换句话说，这三个例子有效 c ++ 0 ？

Or in other words are these three examples valid c++0?

struct Foo
{
   // 1
   int bar( int ) = delete;
};


// 2
int baz( int ) = delete;


template< typename T >
int boo( T t );

// 3
template<>
int boo<int>(int t) = delete;

推荐答案

C ++ 0x spec（& .fct.def.delete]）不拒绝这样的结构，而g ++ 4.5识别所有这3个。

The C++0x spec (§[dcl.fct.def.delete]) doesn't deny such constructs, and g++ 4.5 recognize all 3 of them.

x.cpp: In function 'int main()':
x.cpp:4:8: error: deleted function 'int Foo::bar(int)'
x.cpp:21:11: error: used here
x.cpp:9:5: error: deleted function 'int baz(int)'
x.cpp:22:2: error: used here
x.cpp:9:5: error: deleted function 'int baz(int)'
x.cpp:22:8: error: used here
x.cpp:17:5: error: deleted function 'int boo(T) [with T = int]'
x.cpp:23:7: error: used here

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