模板参数默认为后一个 [英] Template parameter default to a later one

问题描述

此链接无法回答我的问题，所以我会在这里问：

This link doesn't answer my question so I'll ask it here:

基本上我想写一个模板函数

Basically I want to write a template function

template <typename Out, typename In>
Out f(In x);

这里我总是需要指定 Out 调用 f 。我不想每次都这样做，所以我基本上想要

Here I always need to specify Out when calling f. I don't want to do it every time, so I basically want

template <typename Out = In, typename In>
Out f(In x);

这意味着如果我不指定 Out ，它将默认为 In 。但是，这在C ++ 11中是不可能的。

Which means if I don't specify Out, it will default to In. However, this is not possible in C++11.

所以我的问题是，是否有任何方式实现效果：

So my question is, is there any way to achieve the effect:

1. 调用 f（t）将实例化 f code>或更常见的 f


2. c> f U（t）将实例化 f

1. calling f(t) will instantiate f<T,T>(t) or more generally f<typename SomeThing<T>::type, T>
2. calling f<U>(t) will instantiate f<U, T>(t)

推荐答案

我有一个PERFECT解决方案！ f< const int&> 将无法工作，因为函数无法返回对临时的引用，与此处使用的技术无关。

I have a PERFECT solution here! f<const int&> won't work because a function can't return a reference to a temporary, not related to the techniques used here.

[hidden]$ cat a.cpp
#include <iostream>
#include <type_traits>
#include <typeinfo>
using namespace std;

template <typename Out, typename In>
Out f_impl(In x) {
  cout << "Out=" << typeid(Out).name() << " " << "In=" << typeid(In).name() << endl;
  return Out();
}

template <typename T, typename... Args>
struct FirstOf {
  typedef T type;
};

template <typename T, typename U>
struct SecondOf {
  typedef U type;
};

template <typename... Args, typename In>
typename enable_if<sizeof...(Args) <= 1, typename FirstOf<Args..., In>::type>::type f(In x) {
  typedef typename FirstOf<Args..., In>::type Out;
  return f_impl<Out, In>(x);
}

template <typename... Args, typename In>
typename enable_if<sizeof...(Args) == 2, typename FirstOf<Args...>::type>::type f(In x) {
  typedef typename FirstOf<Args...>::type Out;
  typedef typename SecondOf<Args...>::type RealIn;
  return f_impl<Out, RealIn>(x);
}

int main() {
  f(1);
  f(1.0);
  f<double>(1);
  f<int>(1.0);
  f<int>(1);
  f<const int>(1);
  f<int, double>(1);
  f<int, int>(1);
  f<double, double>(1);
}
[hidden]$ g++ -std=c++11 a.cpp
[hidden]$ ./a.out
Out=i In=i
Out=d In=d
Out=d In=i
Out=i In=d
Out=i In=i
Out=i In=i
Out=i In=d
Out=i In=i
Out=d In=d

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