如何从字符串转换为 char*

Question

I have this code, in which I need to input some data, and then the program needs to put the data in alphabetical order.

The problem is that I can't convert string name to the char* variables s1 and s2 .

Data to input:

1 Cheile_Dobrogei 45 25 200
2 Vulcanii_Noroiosi 46 25 50
3 Cetatea_Istra 45 25 100
4 Barajul_Siriu 51 30 50
5 Castelul_Peles 45 30 150
6 Castelul_Bran 53 30 150
7 Voronet 54 35 200
8 Cheile_Bicazului 55 35 100
9 Manastirea_Varatec 56 35 50

#include <iostream>
#include <string>

using namespace std;

struct obiectiv {
    int id;
    string name;
    double latitud; 
    double longitud; 
    double cost_vizitare;
};

int main()
{
    int i, k, temp;
    struct obiectiv ob[9];
    cout << "Introduceti obiectivele(maxim 9): ID NAME LATITUD LONGITUD PRICE" << endl;
    for (i = 0; i < 9; i++) {
        cin >> ob[i].id >> ob[i].name >> ob[i].latitud >> ob[i].longitud >> ob[i].cost_vizitare;
    }

    struct obiectiv tempob[9];
    struct obiectiv t[9];
    for (i = 0;i < 9;i++) {
        tempob[i] = ob[i];
    }
    int sorted;
    for (k = 0; k < 9;k++) {
        sorted = 1;
        for (i = 0;i < 9;i++) {
            char* s1 = tempob[i].name;
            char* s2 = tempob[i + 1].name;
            if (strcmp(s1,s2) > 0) {
                t[i] = ob[i];
                tempob[i] = tempob[i + 1];
                tempob[i + 1] = t[i];
                sorted = 0;
            }
        }
        if (sorted == 1) {
            break;
        }
    }
    cout << "alphabetical order: ";
    for (i = 0; i < 9; i++) {
        cout << tempob[i].name << endl;
    }
}

Answer 1

It's not necessary to even use C-strings in your code. Change 3 lines

char* s1 = ...;
char* s2 = ...;

to

const std::string &s1 = ...;
const std::string &s2 = ...;

and

if (strcmp(s1,s2) > 0)

to

if (s1 > s2) {

and you can keep std::string :

#include <array>
#include <iostream>
#include <string>

struct obiectiv {
    int id;
    std::string name;
    double latitud; 
    double longitud; 
    double cost_vizitare;
};

int main()
{
    std::array<obiectiv, 9> ob;
    std::cout << "Introduceti obiectivele(maxim 9): ID NAME LATITUD LONGITUD PRICE" << '\n';
    for (int i = 0; i < 9; i++) {
        std::cin >> ob[i].id >> ob[i].name >> ob[i].latitud >> ob[i].longitud >> ob[i].cost_vizitare;
    }

    auto tempob = ob;
    
    for (int k = 0; k < 9;k++) {
        bool sorted = true;
        for (int i = 0; i < 9; i++) {
            const auto &s1 = tempob[i].name;
            const auto &s2 = tempob[i + 1].name;
            if (s1 > s2) {
                std::swap(tempob[i], tempob[i + 1]);
                sorted = false;
            }
        }
        if (sorted) {
            break;
        }
    }
    std::cout << "alphabetical order: ";
    for (int i = 0; i < 9; i++) {
        std::cout << tempob[i].name << '\n';
    }
}

There was an error in the swap logic. I replaced it with std::swap to fix the error.

Using std::array instead of C-arrays allows you to copy an array without loop.

I also changed some bad styles in the code. I removed using namespace std; , struct in front of a struct initialization, unnecessary std::endl and I made sorted a boolean.

Answer 2

It's possible to do this(convert from string to char*) in many different ways:

1.Use const_cast operator:

std::string str = "from string to char*";
char *chr = const_cast<char*>(str.c_str());
std::cout << chr << "\n";

2.Use strcpy():

std::string str = "from string to char*";
char *chr = strcpy(new char[str.length() + 1],str.c_str());
std::cout << chr << "\n";

3.Use copy():

std::string str = "from string to char*";
int length = str.size();
char *chr = new char[length + 1];
std::copy(str.begin(),str.end(),chr);
chr[length] = "\0";//add end of line
std::cout << chr << "\n";
delete[] chr; //don't forget!

4.Use contigous storage of std::string:

std::string str = "from string to char*";
char *chr = &*str.begin();
std::cout << chr << "\n";