How to convert string to C-style string

Question

I want to read a config file using C++

My code is here:

myutils.h

#include <string>
#include <map>
using namespace std;

void read_login_data(char *login_data,map<string,string> &data_map); 

myutils.cpp

#include <fstream>
#include <string>
#include <map>
#include "myutils.h"

using namespace std;

void read_login_data(char *login_data,map<string,string> &data_map)
{
    ifstream infile;
    string config_line;
    infile.open(login_data);
    if (!infile.is_open())
    {
        cout << "can not open login_data";
        return false;

    }
    stringstream sem;
    sem << infile.rdbuf();
    while(true)
    {
        sem >> config_line;
        while(config_line)
        {
            size_t pos = config_line.find('=');
            if(pos == npos) continue;
            string key = config_line.substr(0,pos);
            string value = config_line.substr(pos+1);
            data_map[key]=value;

        }
    }


}

test.cpp:

#include <iostream>
#include <map>
#include "myutils.h"

using namespace std;

int main()
{
    char login[] = "login.ini";
    map <string,string> data_map;

    read_login_data(login,data_map);
    cout<< data_map["BROKER_ID"]<<endl;
    char FRONT_ADDR[20]=data_map["BROKER_ID"].c_str();
    cout << FRONT_ADDR<<endl;
}

The config file is:

BROKER_ID=66666
INVESTOR_ID=00017001033

When I compile it using g++ -o test test.cpp , the output is:

young001@server6:~/ctp/ctp_github/trader/src$ g++ -Wall -o test test.cpp   
test.cpp: In function ‘int main()’:  
test.cpp:23:50: error: array must be initialized with a brace-enclosed initializer

how can I assign the data_map["BROKER_ID"] to FRONT_ADDR ?

I have used

strncpy(FRONT_ADDR, data_map["BROKER_ID"].c_str(), sizeof(FRONT_ADDR));

But when I compile it, it says:

young001@server6:~/ctp/ctp_github/trader/src$ g++ -Wall -o test test.cpp   
/tmp/cc5iIQ6k.o: In function `main':  
test.cpp:(.text+0x70): undefined reference to `read_login_data(char*, std::map<std::basic_string<char, std::char_traits<char>, std::allocator<char> >, std::basic_string<char, std::char_traits<char>, std::allocator<char> >, std::less<std::basic_string<char, std::char_traits<char>, std::allocator<char> > >, std::allocator<std::pair<std::basic_string<char, std::char_traits<char>, std::allocator<char> > const, std::basic_string<char, std::char_traits<char>, std::allocator<char> > > > >&)'  
collect2: ld returned 1 exit status

Answer 1

char FRONT_ADDR[20]=data_map["BROKER_ID"].c_str();

You can't do this. You should use strncpy (or copy_n , but in this case you should check, that length of data_map["BROKER_ID"].c_str() is greater or equal to n ) to copy one array of char into another.

std::strncpy(FRONT_ADDR, data_map["BROKER_ID"].c_str(), sizeof(FRONT_ADDR));

And, the best choose, of course, using std::string :

std::string FRONT_ADDR = data_map["BROKER_ID"];
// and, anywhere you need const char*
somefunction(FRONT_ADDR.c_str());

Answer 2

Try this,

char FRONT_ADDR[20]={0}; //all elements 0
std::strcpy(FRONT_ADDR, data_map["BROKER_ID"].c_str()); // assuming there is enough space in FRONT_ADDR

Answer 3

In C++, there is an inbuilt method( c_str() ) to do the conversion. Here's an example.

char* cs[10];
Qstring s("string");
cs = s.c_str();