如何“返回值"?的运算符>>流课程的工作? [英] How does the "return value" of operator >> of istream class work?

问题描述

我试图理解以下说法:

    int main() {
        fstream inf( "ex.txt", ios::in );
        char c;
        while( inf >> c ) {
            cout << c << ", ";
        }
        return 0;
    }

( inf>> c )在上面的while循环中返回什么?我下载了gcc源代码并尝试使用它，但是它对我来说太复杂了:(.

What does ( inf >> c ) return in the while loop above? I downloaded the gcc source code and tried to play around with it but it was way too complex for me :( .

我检查了C ++参考页，发现它返回了对自身的引用:

I checked the C++ reference page, I realized that it return a reference to itself:

    istream& operator>> (bool& val );
    istream& operator>> (short& val );
    istream& operator>> (unsigned short& val );
    istream& operator>> (int& val );
    istream& operator>> (unsigned int& val );
    istream& operator>> (long& val );
    istream& operator>> (unsigned long& val );
    istream& operator>> (float& val );
    istream& operator>> (double& val );
    istream& operator>> (long double& val );
    istream& operator>> (void*& val );

    istream& operator>> (streambuf* sb );

    istream& operator>> (istream& ( *pf )(istream&));
    istream& operator>> (ios& ( *pf )(ios&));
    istream& operator>> (ios_base& ( *pf )(ios_base&));

    *** the following functions are not members but GLOBAL functions:

    istream& operator>> (istream& is, char& ch );
    istream& operator>> (istream& is, signed char& ch );
    istream& operator>> (istream& is, unsigned char& ch );

    istream& operator>> (istream& is, char* str );
    istream& operator>> (istream& is, signed char* str );
    istream& operator>> (istream& is, unsigned char* str );

所以我创建了一个类似的类，比方说my_istream:

So I created a similar class, let's say my_istream:

struct my_istream {
    my_istream& self_ref;
};

int main() {
    my_istream mis;
}

编译时出现此错误:

1>c:\users\chan\documents\visual studio 2010\projects\topcoder\topcoder\main.cpp(26): error C2758: 'my_istream::self_ref' : must be initialized in constructor base/member initializer list

但是，我真的不知道在这种情况下应将self_ref初始化为什么?在处理链表时，我了解指向自身的指针，并且我了解C ++中的reference(& )只是C中伪装的指针形式.但是我无法解释这种情况吗? istream 的内部实现实际上是如何工作的?引用如何被评估为真或假?谢谢！

However, I don't really know what should self_ref be initialized to in this case? I understand a pointer to itself when dealing with linked-list, and I understand that reference( & ) in C++ is just a disguised form of pointer in C. But I can't explain this situation? How does the internal implementation of istream actually work? How does a reference can be evaluated as true or false? Thanks!

struct my_istream {
    my_istream() {
    }

    my_istream& operator >>( int x ) {
        return *this;
    }
};

int main() {
    my_istream mis;
    int x;
    while( mis.operator>>( x ) ) {
        cout << "--";
    }
}

我应该在my_istream类中添加什么才能在while循环中工作?

What should I add to my_istream class in order to work inside the while loop?

推荐答案

要返回自引用，只需 return * this .您实际上并没有声明内部自引用.您可以 具有引用成员，但是必须在构造函数初始化器列表中对其进行初始化: class Foo {Foo(int& some_int_ref):my_ref_member(some_int_ref)

To return a self ref you just return *this. You don't actually declare an internal self reference. You can have reference members, but you must initialize them in the constructor initializer list: class Foo { Foo(int& some_int_ref) :my_ref_member(some_int_ref)

istream还有另一个重载来确定其布尔值.

istream has another overload to determine its boolean value.

可链接成员的示例(例如 operator>> ，但在这种情况下只是一个普通函数)，并给一个类一个布尔值(尽管后者是一个复杂的主题，值得其自己的主题):

Example of a chainable member (like operator >>, but in this case just an ordinary function) and giving a class a boolean value (although the latter is a complex subject worthy of its own subject):

#include <iostream>

class X {
    bool B;
public:
    X() :B(false) { }
    X& toggle() { B = !B; return *this; }
    operator void*() { return B ? this : 0; }
};

int main()
{
    X x;

    x.toggle().toggle().toggle();
    if (x)
        std::cout << "true!" << std::endl;
}

我不想在此答案中深入研究 operator bool 与 operator void * ，但是这个古老的stackoverflow问题应该为您提供良好的参考:href ="https://stackoverflow.com/questions/2145931/why-is-operator-bool-invoked-when-i-cast-to-long">为什么"operator bool()"当我强制转换为"long"时调用?

I didn't want to delve into operator bool vs operator void * in this answer, but this old stackoverflow question should give you good references: Why is "operator bool()" invoked when I cast to "long"?

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