c++ - 为什么这个 std::bind 没有转换为 std::function？

Question

为什么嵌套std::bind在下面的代码中没有隐式转换为 std::function<void()>任何主要编译器(VS2010/2012、gcc、clang)？这是标准行为还是错误？

#include <functional>

void bar(int, std::function<void()>) { }
void foo() { }

int main()
{
    std::function<void(int, std::function<void()>)> func;
    func = std::bind(bar, 5, std::bind(foo));

    std::cin.get();
    return 0;
}

Answer 1

这在 boost documentation 中有解释:

The inner bind expressions are evaluated, in unspecified order, before the outer bind when the function object is called; the results of the evaluation are then substituted in their place when the outer bind is evaluated. In the example above, when the function object is called with the argument list (x), bind(g, _1)(x) is evaluated first, yielding g(x), and then bind(f, g(x))(x) is evaluated, yielding the final result f(g(x)).

Boost 甚至提供了 protect 来防止这种评估:

#include <boost/bind/protect.hpp>
...
func = std::bind(bar, 5, boost::protect(std::bind(foo)));

但是，要调用 func，您必须像这样提供两个参数(感谢 David Rodríguez - dribeas 指出这一点)，所以这个例子绝对不好:

func(1, std::function<void()>());