對於運算符“ <<”的歧義過載

Question

我正在學習C ++中的運算符重載，我想知道以下代碼的輸出

#include<iostream>
using namespace std;
class xyz
{

 public:
        int i;

    friend ostream & operator<<( ostream & Out , int);
};



ostream & operator<<(ostream & out , int i)
{
    cout<<10+i<<endl;
}


int main()
{
    xyz A;
    A.i=10;

    cout<<10;
}

我有兩個錯誤

1. 錯誤：'operator <<'的模棱兩可的重載（操作數類型為'std :: ostream {aka std :: basic_ostream}'和'int'）cout << 10 + i;

2. 錯誤：'operator <<'的模棱兩可的重載（操作數類型為'std :: ostream {aka std :: basic_ostream}'和'int'）cout << 10;

誰能解釋這個問題是什么？

我想知道如果我重載“ <<”運算符以僅使用一個參數int（obvious）打印int，而我只想單獨打印一個數字，例如上述代碼“ cout << 10” int，會發生什么。 因此，當我嘗試僅打印任何整數時，編譯器將如何決定應調用哪個函數。

Answer 1

所以很明顯，問題是您已經編寫了ostream & operator<<(ostream & out , int i) 。 但是很明顯，您要寫的是這樣的

ostream& operator<<(ostream& out, const xyz& a) // overload for xyz not int
{
    out<<a.i<<endl; // use out not cout
    return out;     // and don't forget to return out as well
}

和這個

int main()
{
    xyz A;
    A.i=10;

    cout<<A<<endl; // output A not 10
}

Answer 2

// this include brings std::ostream& operator<<(std::ostream&, int)
// into scope and therefore you cannot define your own later
#include<iostream>  

using namespace std;
class xyz
{

 public:
        int i;

    // needs body 
    friend ostream & operator<<( ostream & Out , int)
    {
        return Out;
    }
};



/* cant have this after including ostream
ostream & operator<<(ostream & out , int i)
{
    cout<<10+i<<endl;
}
*/


int main()
{
    xyz A;
    A.i=10;

    cout<<10;
}