是否可以在模板中键入指向外部"C"函数类型的指针? [英] Is it possible to typedef a pointer-to-extern-"C"-function type within a template?

问题描述

我想在模板中添加公共typedef，以指向使用"C"语言链接的函数接受一参数的指针.

I want to add a public typedef to a template for a pointer to a function-taking-one-argument that uses "C" language linkage.

我尝试过:

extern "C" {
    template <typename return_t_, typename arg1_t_>
    struct test
    {
        typedef return_t_ (*C_fun1_t)(arg1_t_);
    };
}

并且:

template <typename return_t_, typename arg1_t_>
struct test
{
    extern "C" {
        typedef return_t_ (*C_fun1_t)(arg1_t_);
    }
};

并且:

template <typename return_t_, typename arg1_t_>
struct test
{
    extern "C" typedef return_t_ (*C_fun1_t)(arg1_t_);
};

没有成功.

我要实现的目标是可能的吗?

Is what I am trying to accomplish possible?

推荐答案

C ++ 03，第7.5p4节:

C++03, §7.5p4:

链接规范应仅在命名空间范围内发生. …类成员名称和类成员函数的成员函数类型将忽略C语言链接.

A linkage-specification shall occur only in namespace scope. … A C language linkage is ignored for the names of class members and the member function type of class member functions.

不幸的是，您在当前的C ++中根本无法做到这一点.该文本在最新的C ++ 0x草案中未更改，但"template typedefs"也许可以完成.

Unfortunately, you simply can't do this in current C++. This text is unchanged in the latest C++0x draft, but "template typedefs" may be able to accomplish it.

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