如何过载操作符<对于C ++中的数组？ [英] how overload operator<< for an array in C++?

问题描述

我试图这样做：

template <typename T>
ostream &operator<<(ostream &os, T &arr)
{ /*...*/ }

但是 T 代表一个数组？为数组重载< 运算符是否正确？

But can T represent an array? Is it correct to overload the << operator for an array?

编辑：

根据Kerrek SB的建议，这里是< / code>：

According to Kerrek SB's advice, here is my implementation for <<:

template <typename T, unsigned int N>
ostream &operator<<(ostream &os, const T (&arr)[N])
{
    int i;
    for(i = 0; i < N; i++)
        os << arr[i] << " ";
    os << endl;
    return os;
}

我的实现是否正确？我遇到了编译错误。

Is my implementation right? I got a compilation error.

推荐答案

您可以这样做：

template <typename T, unsigned int N>
std::ostream & operator<<(std::ostream & os, const T (&arr)[N])
{
  // ..
  return os;
}

这当然只适用于编译时数组。请注意， T 是内置类型或 std 中的类型时，不允许实例化此模板，命名空间！

This works only for compile-time arrays, of course. Note that you are not allowed to instantiate this template when T is a built-in type or a type in the std namespace!

如果可能的话，最好尽可能使这个内联，因为你会为每个 N 。 （漂亮的打印机有此示例。）

Probably best to make this inline if possible, since you'll cause a separate instantiation for every N. (The pretty printer has an example of this.)

你会注意到，毯子模板引入了歧义，因为 os<< Hello现在有两个可能的重载：模板匹配 const char（&）[6] 对于衰减到指针 const char * ，它们都具有相同的转换序列。我们可以通过禁用字符数组的重载来解决这个问题：

You will notice, though, that the blanket template introduces an ambiguity, because os << "Hello" now has two possible overloads: the template matching const char (&)[6], and the (non-template) overload for the decay-to-pointer const char *, which both have identical conversion sequences. We can resolve this by disabling our overload for char arrays:

#include <ostream>
#include <type_traits>

template <typename T, unsigned int N>
typename std::enable_if<!std::is_same<T, char>::value, std::ostream &>::type
operator<<(std::ostream & os, const T (&arr)[N])
{
  // ..
  return os;
}

事实上，为了更加通用，你还可以使 basic_ostream 参数模板参数：

In fact, to be even more general you can also make the basic_ostream parameters template parameters:

template <typename T, unsigned int N, typename CTy, typename CTr>
typename std::enable_if<!std::is_same<T, char>::value,
                        std::basic_ostream<CTy, CTr> &>::type
operator<<(std::basic_ostream<CTy, CTr> & os, const T (&arr)[N])
{
  // ..
  return os;
}

鉴于 T 必须是用户定义的类型，您甚至可以用 is_fundamental< T> is_same $ c>以获得更多的检查（但用户仍然不能使用这个标准库类型的数组）。

In view of the fact that T must be a user-defined type, you could even replace is_same<T, char> with is_fundamental<T> to get a bit more checking (but users still must not use this for arrays of standard library types).

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