设置std :: function变量来引用std :: sin函数

Question

我有一个关于如何正确使用新的C++ 11 std::function变量的问题.我已经看过几个搜索互联网的例子,但它们似乎并没有涵盖我正在考虑的用例.拿这个最小的例子,其中函数fdiff是定义的有限前向差分算法的实现numerical.hxx(这不是问题,我只是想给出一个上下文的原因,为什么我想要采用任意函数并传递它) .

#include <functional>
#include <iostream>
#include <cmath>
#include "numerical.hxx"

int main()
{
    double start = 0.785398163;
    double step  = 0.1;
    int    order = 2;

    std::function<double(double)> f_sin = std::sin;

    std::cout << fdiff(start, step, order, f_sin) << std::endl;

    return 0;
}

试图编译上面的程序给我错误(在clang ++中)

test.cpp:11:32: error: no viable conversion from '<overloaded function type>' to
      'std::function<double (double)>'
        std::function<double(double)> f_sin = std::sin;
                                      ^       ~~~~~~~~
/usr/lib/gcc/x86_64-unknown-linux-gnu/4.7.1/../../../../include/c++/4.7.1/functional:2048:7: note: 
      candidate constructor not viable: no overload of 'sin' matching
      'nullptr_t' for 1st argument
      function(nullptr_t) noexcept
      ^
/usr/lib/gcc/x86_64-unknown-linux-gnu/4.7.1/../../../../include/c++/4.7.1/functional:2059:7: note: 
      candidate constructor not viable: no overload of 'sin' matching 'const
      std::function<double (double)> &' for 1st argument
      function(const function& __x);
      ^
/usr/lib/gcc/x86_64-unknown-linux-gnu/4.7.1/../../../../include/c++/4.7.1/functional:2068:7: note: 
      candidate constructor not viable: no overload of 'sin' matching
      'std::function<double (double)> &&' for 1st argument
      function(function&& __x) : _Function_base()
      ^
/usr/lib/gcc/x86_64-unknown-linux-gnu/4.7.1/../../../../include/c++/4.7.1/functional:2092:2: note: 
      candidate template ignored: couldn't infer template argument '_Functor'
        function(_Functor __f,
        ^
1 error generated.

或者来自g ++

test.cpp: In function ‘int main()’:
test.cpp:11:45: error: conversion from ‘<unresolved overloaded function type>’ to non-scalar type ‘std::function<double(double)>’ requested

正如我理解这个问题,它是因为它std::sin是作为标准库中的模板类实现的,但我似乎无法弄清楚我需要做些什么来提供足够的特化来获取函数引用.我也尝试了各种各样的东西,比如使用new auto关键字,使用&std::sin指针等,但它们都给了我相同类型的错误.

Answer 1

std::sin是一个重载函数:你必须消除std::sin你的意思:

std::function<double(double)> f_sin = (double(*)(double))&std::sin;

在某些情况下,编译器可以消除重载函数的歧义(例如,如果f_sin是类型double(*)(double),则不需要强制转换).但是,这不是其中之一.