重载的运算符& lt;& lt;流级联问题

问题描述

我有以下代码:

struct simple
{
    simple (int a1, int a2) : member1(a1), member2(a2) {}
    int member1;
    int member2;
};

std::ofstream &operator << (std::ofstream &f, const simple &obj)
{
    f<<obj.member1<<", "<<obj.member2;
    return f;
}
int main(int argc, const char *argv[])
{
    std::ofstream f("streamout.txt");

    simple s(7,5);
    f << s;               //#1 This works
    f << "label: " << s;  //#2 This fails

    return 0;
}

我试图理解为什么#1可以工作，而在尝试使用重载运算符将其像#2那样进行连接时却出现了问题，但出现以下错误(在MacOSX上为gcc 4.5.3):

I'm trying to understand why #1 works, while there are problems when trying to use the overloaded operator concatenating it as in #2 which fails with the following error (gcc 4.5.3 on MacOSX):

如果我将运算符定义为

std::ostream &operator << (std::ostream &f, const simple &obj)
{ ... }

类似于过载解析的声音，在ofstream中插入已经提供了过载的东西(在这种情况下为const char *"label")，在过载解析后会破裂，但是我不能真正做到这一点.了解这里到底发生了什么.我想清楚地了解编译器正在尝试执行的操作.

Sounds like something related to overload resolution, where having a something inserted in the ofstream for which there's already a provided overload (the const char * "label" in this case) breaks up following overload resolution, but I can't really understand what exactly is going on here.I'd like to get a clear picture of what the compiler's trying to do..

推荐答案

在线:

f << "label: " << s;

由于第一次调用 operator<< 会返回 std :: ostream& ，因此第二次编译失败:运算符的左操作数不是再输入 std :: ofstream ，就不会找到您的重载.

Because the first call to operator<< returns a std::ostream &, the second fails to compile : the left operand to the operator is not of type std::ofstream anymore and your overload is not found.

您应该真正使用第二个签名，因为我认为没有理由将您的类型限制为输出到 std :: ofstream .

You should really use the second signature, as I see no reason for restricting your type to be outputted to std::ofstream.

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