c++ - 如何将字符串转换为C风格的字符串

Question

我想使用 C++ 读取配置文件

我的代码在这里:

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;

        }
    }


}

测试.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;
}

配置文件是:

BROKER_ID=66666
INVESTOR_ID=00017001033

当我使用 g++ -o test test.cpp 编译它时，输出是:

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

如何将 data_map["BROKER_ID"] 分配给 FRONT_ADDR？

我用过

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

但是当我编译它时，它说:

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();

你不能这样做。您应该使用 strncpy(或 copy_n，但在这种情况下您应该检查 data_map["BROKER_ID"].c_str() 的长度大于或等于 n) 将一个 char 数组复制到另一个。

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

当然，最好的选择是使用 std::string:

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