模板化的成员函数typedefs无法编译 [英] Templated member function typedefs won't compile

问题描述

#include <iostream>
#include <string>
using namespace std;

void printstr( const string & s ) { cout << s << endl; }

template < typename A >
class Test
{
public:
    typedef void (*Func)( const A & );
};

typedef void (*Func)( const string & );

template < typename A >
void bind(
        Test< A >::Func f,           //<---- does NOT compile
        //Func f,                    //<---- compiles & works!
        //void (*f)( const A & ),    //<---- compiles & works!
        const A & a) { f( a ); }


int main( )
{
    bind( printstr, string("test") );
    return 0;
}

在上面的代码中，我试图使用另一个类的函数指针typedef.如图所示，它不会编译，但是其他两行中的任何一条都未注释，而不是 Test<.> :: Func f，行，编译良好！这是我在C ++中做不到的事情吗?需要什么语法?

In the code above, I am trying to use a function pointer typedef from another class. As shown, it does not compile, but with either of the other two lines uncommented instead of the Test< A >::Func f, line, it compiles fine! Is this something I can't do in C++? What syntax is needed?

使用g ++ 4.4.3，我得到

Using g++ 4.4.3, I get

test.cpp:20: error: variable or field "bind" declared void
test.cpp:20: error: expected ")" before "f"
test.cpp:23: error: expected primary-expression before "const"

推荐答案

名称 Test< A> :: Func 是从属名称，需要以 typename

The name Test<A>::Func is a dependent name and needs to be prefixed with typename

typename Test< A >::Func f,  

要获得更详细的解释，您应该在以下答案中查看约翰内斯的解释

For a more detailed explanation you should check out Johannes explanation in the following answer

• 在哪里以及为什么我必须将模板"放在和类型名称"关键字?

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