CODEWITHSUNDEEP
c++visual-studioclassvisual-studio-2017operator-overloading
Object Unknown
Object Unknown

Reputation: 315

How to fix "There too many argument in this opreator"

I build a new string class in C++ called UString, I think it should support cin and cout, so I want to overload << and >> operator, but my Visual Studio told me that I put too many argument in that function.

What should I do?

My first code like this:

std::istream &operator>>(UString ustr)
{
    std::istream istr(nullptr);
    istr >> ustr.str;
    return istr;
}

I didn't got any error if I compile the class without use this function,but if I use it by: cin>>ustr;, the compiler told me that there is no match ver of this >> operator. So I searched it in Google and changed it like that:

std::ostream &operator<< (std::ostream& ostr, const UString ustr)
{
    ostr << ustr.str;
    return ostr;
}

But, the pre-check told me that there are too many arguments in it, it can not be compiled successfully also.

My IDE is Visual Studio 2017, C++ ver is set to C++17.

std::ostream &operator<< (std::ostream& ostr, const UString ustr)
{
    ostr << ustr.str;
    return ostr;
}
//>>
std::istream &operator>>(UString ustr)
{
    std::istream istr(nullptr);
    istr >> ustr.str;
    return istr;
}

After overloading, it should support use like this:

UString ustr;
cin>>ustr;
cout<<ustr<<endl;

Upvotes: 0

Views: 92

Answers (1)

Serikov
Serikov

Reputation: 1209

std::istream&>> and << operators are binary operators (they have two arguments). When you overload this operators you can make it either member function (inside your class) or free function.

Code x >> y; will be transformed either to x.operator>>(y); (in case of member function) or to operator>>(x, y); (in case of free function).

As you want to overload operator with left argument of type that you can't modify (stream) you should overload it as a free function.

class UString { ... };

std::istream& operator>>(std::istream& s, UString& str) { ... }

std::ostream& operator<<(std::ostream& s, Ustring& str) { ... }

Upvotes: 1

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