C2679 ошибка вижуал студия

Please don’t confuse with the title as it was already asked by someone but for a different context

The below code in Visual C++ Compiler (VS2008) does not get compiled, instead it throws this exception:

std::ifstream input (fileName);   

while (input) {
  string s;
  input >> s;
  std::cout << s << std::endl;
};

But this code compiles fine in cygwin g++. Any thoughts?

Adding to @sbi answer, in my case the difference was including <string> instead of <string.h> (under VS 2017).

See the following answer: similar case answer

In addition to what others said. The following code was necessary in my application to compile succesfully.

std::cout << s.c_str() << std::endl;

Another work-around to this is go to project properties -> General -> Character Set and choose «Ues Multi-Byte Character Set» (You won’t need to use c_str() to output the string)

There’s disadvantages to using MBCS so if you plan to localize your software, I’d advize against this.

answered Oct 24, 2018 at 16:10

HI. Before my question, I’ve looked in other threads with same error, nothing worked for me.
My problem code is at in >> a.name >> a.extension;. As I tested myself, if I change the type from string to char for my variables, it will work, but I can’t make it work with a string type of value.

Am I doing something wrong?
Below full error code on compillation (Visual Studio 2015)

Error C2679 binary ‘>>’: no operator found which takes a right-hand
operand of type ‘const std::string’ (or there is no acceptable
conversion)

Thanks in advance.

#include <iostream>
#include <ctime>
#include <string>
using namespace std;

class Soft {
private:
    int *size;
    time_t *dateTime;
public:
    string name;
    string extension;
    Soft();
    Soft(string, string);
    Soft(const Soft&source);
    ~Soft();

    friend istream& operator>> (istream& in, const Soft& a) 
    {
        in >> a.name >> a.extension;
        return in;
    };

    void changeName(string);
    void changeExtension(string);
};

Soft::Soft() {
    size = new int;
    *size = 0;
    name = string("");
    extension = string("");
    dateTime = new time_t;
    *dateTime = time(nullptr);
}

Soft::Soft(const Soft&source) {
    name = source.name;
    extension = source.extension;
    size = new int;
    *size = *source.size;
    dateTime = new time_t;
    *dateTime = time(nullptr);
}

Soft::Soft(string a, string b) {
    name = a;
    extension = b;
    dateTime = new time_t;
    *dateTime = time(nullptr);
}

Soft::~Soft() {
    delete[] size;
    delete[] dateTime;
}

void Soft::changeExtension(string a) {
    extension = a;
}
void Soft::changeName(string a) {
    name = a;
}


int main() {

    Soft a;

    getchar();
    getchar();
    return 0;
}

asked Oct 1, 2016 at 14:24

Assignment operator’s l-Value mustn’t be constant because this operator changes the values so trying to change a constant value is a trying to break the rules.

look at insertion operator:

ostream& operator <<(ostream&, const myCalss&); // here const is ok

here it’s ok because the insertion operator is just used to print values (const and non-const) without changing them.

*** if your really need to Break the rules of constness declare your member data as mutable:

     mutable int *size;
     mutable time_t *dateTime;

but in your example you needn’t to do so, I just explain that you can change cont variable.

answered Oct 1, 2016 at 15:05