c++ function with variable number of arguments perform wrong 'string'-type convertion

Question

I ran into such a problem: i wrote some code and when i try to format some data (especially strings) i got strange result

#include <iostream>     // std::cout, std::endl
#include <string.h>
#include <vector>
#include <cstdarg>
#include <sstream>

using namespace std;

void va_cout(string format, ...)
{
    vector<char> buf(256);
    va_list args;
    va_start(args, format);
    vsnprintf_s(&buf[0], buf.size(), buf.size() + strlen(format.c_str()), format.c_str(), args);
    va_end(args);
    cout << endl << "apply va_cout(""%s"", s) = " << &buf[0] << endl;
}

int main() {
    string s("myString");
    cout << endl << s << endl;
    va_cout("%s", s);
    va_cout("%s", s.c_str());
    return 0;
}

// results

myString
apply va_cout(%s, s) = └¤#
apply va_cout(%s, s) = myString

why it returns line 2 ?

Remy Lebeau · Accepted Answer · 2017-04-29 07:07:13Z

4

%s can be used only with char * and const char *. You are passing a std::string and treating it like a char * in vsnprintf_s(). There is no implicit conversion from std::string to char *. You need to call std::string::c_str() for that.

edited Apr 29, 2017 at 7:07

Remy Lebeau

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answered Apr 29, 2017 at 6:06

Martin Perry

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1 Comment

Nikolay Lavnikevich Over a year ago

thank you. i'm trying to understand how to round out std::string::c_str() using.

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