CODEWITHSUNDEEP
c++
Zodrak
Zodrak

Reputation: 49

error C2678: binary '>>' : no operator found which takes a left-hand operand of type 'std::basic_istream<_Elem,_Traits>'

I get this strange error. I think I have included all necessary files. What could cause this? The error is caused by this line :

cin >> x >> "\n" >> y >> "\n";

Here is the code:

#ifndef ARITMETICE_H
#define ARITMETICE_H

#include <iostream>
#include <string>

#include "UI\comanda.h"

using namespace Calculator::Calcule;

namespace Calculator{
namespace UI{

    template<int Operatie(int, int)>
    class CmdAritmetice : public ComandaCalcule
    {

    public:
        CmdAritmetice(const string &nume) : ComandaCalcule(nume)
        {
        }

        void Execute()
        {
            cout << Nume() << "\n";
            cout << "Introduceti doua numere intregi (x, y)\n";
            int x, y;
            cin >> x >> "\n" >> y >> "\n";   // here a get the error
            cout << x << " " << Nume() << " " << y << " = " << Operatie (x,y) <<"\n";
        }
    };
}
}
#endif

Upvotes: 0

Views: 1602

Answers (5)

user2249683
user2249683

Reputation:

If your intention is to consume white spaces and one new line character (note the are flavors of new line representations), you might write a manipulator

#include <cctype>
#include <iostream>
#include <sstream>

std::istream& nl(std::istream& is) {
    typedef std::istream::traits_type traits;
    while(true) {
        traits::int_type ch;
        if(std::isspace(ch = is.get())) {
            if(ch == '\n') break;
        }
        else {
            is.putback(ch);
            // No space and no '\n'
            is.setstate(std::ios_base::failbit);
            break;
        }
    }
    return is;
}

int main()
{
    std::istringstream s("1\n2");
    int x, y;
    s >> x >> nl >> y >> nl;
    std::cout << x << ' ' << y << '\n';
    if(s.fail()) {
        // The missing '\n' after '2'
        std::cout <<"Failure\n";
    }
    return 0;
}

Upvotes: 0

Ajay
Ajay

Reputation: 18411

cin is object of type istream. This class has overloaded operator >> to take input from console and put the value into given variable. The variable must be l-value, not r-value. In short, the expression given on right of >> must be writable variable.

This will not work:

const int x;
cin >> x;

Simply because x is a const int&, and not int& which what istream::operator>>(int&) expects.

Going this way further, when you make a call:

cin >> "\n";

You are essentially calling operator >> (const char*) and not operator >> (char*), and hence the error. Because of multiple overloads of operator>> and template code behind, the error is not clear.

Note: The exact signature of operator >> might differ, but problem is with const-ness.

Upvotes: 0

bowman han
bowman han

Reputation: 1135

cin >> must has a writable variable in the right, your cin>>"\n" redirect cin to a const char* type, which is read only

in a word, just use cin>> x >> y; and the iostream will do the rest for you.

Upvotes: 0

Lightness Races in Orbit
Lightness Races in Orbit

Reputation: 385088

Not sure what you expected to happen when you tried to extract data from a stream into a string literal!

Upvotes: 1

Igor Tandetnik
Igor Tandetnik

Reputation: 52461

The problem is cin >> "\n". It purports to read user input into a string literal, which doesn't make any sense. Just drop it, make it cin >> x >> y;

Upvotes: 3

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