std::function<R(Args...)>::target
来自cppreference.com
< cpp | utility | functional | function
| template< class T > T* target() noexcept; |
(1) | (C++11 起) |
| template< class T > const T* target() const noexcept; |
(2) | (C++11 起) |
返回指向存储的可调用函数目标的指针。
参数
（无）
返回值
若 target_type() == typeid(T) 则为指向存储的函数的指针，否则为空指针。
示例
运行此代码
#include <functional> #include <iostream> int f(int, int) { return 1; } int g(int, int) { return 2; } void test(std::function<int(int, int)> const& arg) { std::cout << "test function: "; if (arg.target<std::plus<int>>()) std::cout << "it is plus\n"; if (arg.target<std::minus<int>>()) std::cout << "it is minus\n"; int (*const* ptr)(int, int) = arg.target<int(*)(int, int)>(); if (ptr && *ptr == f) std::cout << "it is the function f\n"; if (ptr && *ptr == g) std::cout << "it is the function g\n"; } int main() { test(std::function<int(int, int)>(std::plus<int>())); test(std::function<int(int, int)>(std::minus<int>())); test(std::function<int(int, int)>(f)); test(std::function<int(int, int)>(g)); }
输出：
test function: it is plus test function: it is minus test function: it is the function f test function: it is the function g
参阅
| 获得 std::function 所存储的目标的typeid (公开成员函数) |
