在传递函数的标识符时，无法检查模板参数是否是您遇到的问题 [英] Having trouble checking whether template parameter is something you get when you pass a function's identifier

问题描述

我要实现以下功能:

template<typename Function, typename... Parameters>
inline void foo(
    const Function&   kernel_function,
    bar_t             bar
    Parameters...     parameters)

{
    static_assert(/* magic */,
        "You can only foo() a function, not values of any other type");
    / * etc. etc. */
}   

，我只需要使用函数的标识符或指向函数的指针来调用它:不能使用lambads或具有operator()或std::function s的方法或类.我应该用什么代替/* magic */?仅使用std::is_function似乎无效.

and I need it to only be called with the identifiers of functions, or with pointers to functions: No lambads or methods or classes with operator() or std::functions. What should I replace /* magic */ with? Just using std::is_function doesn't seem to work.

推荐答案

在 ，我们有 std::is_function ，如果返回true您传递给它一个实际的函数和带有lambda的false，带有重载的operator()的类以及指向函数的指针.我们也有 std::is_pointer 和

In <type_traits> we have std::is_function which returns true if you pass it an actual function and false with lambdas, classes with overloaded operator() and pointers to functions. We also have std::is_pointer and std::remove_pointer which can be used to check for and remove the pointer type from a function pointer to test the pointer with std::is_function. Using those your assert would look like

static_assert(std::is_function<Function>::value || 
    (std::is_pointer<Function>::value &&
    std::is_function<std::remove_pointer<Function>::type>::value),
    "You can only foo() a function, not values of any other type");

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