为什么我们需要在重载时返回对 istream/ostream 的引用 >>并且<<运营商?

问题描述

如果我不返回 din 或 dout 会发生什么，实际上我正在读一本书，其中作者返回了流引用

What happens if I do not return din or dout, actually I'm reading a book in which writer returns back stream references

istream & operator>>(istream &din,vector &a)
{
    for(int i=0;i<size;i++)
    din>>a.v[i];
    return din;
}

ostream & operator<<(ostream &dout,vector &a)
{
    dout<<"("<<a.v[0];
    for(int i=1;i<size;i++)
    dout<<", "<<a.v[i];
    dout<<")";
    return dout;
}

推荐答案

原因是几个事实的结合.

The reason is a combination of several facts.

1. 您希望能够像这样链接输入和输出操作

1. You want to be able to chain input and output operations as in

in  >> x >> y;

out << z << std::precision(10) << t << std::endl;

所以你必须返回一些允许 operator<< 的东西.

so you must return something that allows operator<< again.

由于您希望操作符处理任何 istream，即从 std::istream 派生的任何对象，您不能定义

Since you want your operator to work on any istream, i.e. any object derived from std::istream, you cannot define

operator<<(istream_type, object);    // take istream by value

因为这仅适用于特定的 istream 类型 istream_type，而不适用于通用的 istream.为此，必须使用多态性，即采用引用或指针(将是指向从 std::istream 派生的类的引用或指针).

since this would only work for the specific istream type istream_type, but not for a generic istream. For that one must use polymorphism, i.e. either take a reference or a pointer (which will be a reference or pointer to a class derived from std::istream).

由于您只有对 istream 的引用，因此您无法返回 istream 对象本身(其类型可能在 operator<<) 但只有你得到的参考.

Since you only have a reference to the istream, you cannot return the istream object itself (which may be of a type not even defined at the point of the definition of operator<<) but only the reference you've got.

可以通过定义 operator<< 一个 template 并按值获取并返回 istream_type 来绕过此限制，但这需要istream 类型有一个复制构造函数，它很可能没有充分的理由.

One could get around this restriction by defining operator<< a template and take and return the istream_type by value, but that requires the istream type to have a copy constructor, which it may well not have for good reasons.

为了激发多态性，原则上可以使用指针(指向流)而不是引用.但是，operator<<(stream*,const char*) 是在 C++ 中不允许(至少一个操作数必须是类或枚举类型).

In order to envoke polymorphism one could, in principle, use pointers (to streams) rather than references. However, operator<<(stream*,const char*) isnot allowed in C++ (at least one operand must be of class or enumeration type).

因此，对于流指针，必须使用函数调用语法，而您又回到了 C 风格的fprintf(stream*, args...).

Thus, with stream pointers one must use function-call syntax and you're back with C-style fprintf(stream*, args...).

此外，指针可以为空或悬空，这实际上是它们的默认状态(在没有初始化器的情况下声明时)，而可以假定引用是有效的(没有初始化器就不能声明).

Moreover, pointers can be null or dangling, which in fact is their default state (when declared without initializer), while a reference can be assumed to be valid (it cannot be declared without initializer).

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