将整数分配给 char * 指针数组中的索引

Question

Redoing my previous question since I didn't provide enough detail.

I have a char pointer array, char* token[100] . Let's say I have a double-digit number, like 33.

How do I assign this int into an index in the token array, so that when I print out that token it will give me 33 and not some sort of ASCII value?

char* token[100];
int num = 33;
//How do I assign num into a specific token index, like for example:
token[1] = num;
//When I print out that token index, I want 33 to be printed out
cout << token[1] << endl; // I want to have 33 be the result. Right now I have '!' as an output

Answer 1

It seems you mean something like the following

#include <iostream>
#include <string>
#include <cstring>

int main() 
{
    char * token[100] = {};
    int num = 33;

    std::string s= std::to_string( num );

    token[1] = new char[s.size() + 1];

    std::strcpy( token[1], s.c_str() );

    std::cout << "token[1] = " << token[1] << '\n';

    delete [] token[1];

    return 0;
}

The program output is

token[1] = 33

If you are not allowed to use C++ containers and functions then the program can look the following way

#include <iostream>
#include <cstdio>
#include <cstring>

int main() 
{
    char * token[100] = {};
    int num = 33;

    char buffer[12];

    std::sprintf( buffer, "%d", num );

    token[1] = new char[std::strlen( buffer ) + 1];

    std::strcpy( token[1], buffer );

    std::cout << "token[1] = " << token[1] << '\n';

    delete [] token[1];

    return 0;
}

Answer 2

I'm convinced, from comments, that you want an array of integer types. If we get further clarification about why this needs to be a char array, I'll update my answer, but from all available information it seems like you really want an integer-type array.

#include <iostream>

int main(int argc, char** argv)
{
    int token[100] = {};
    int num = 33;

    token[1] = num;

    std::cout << token[1] << std::endl;

    return 0;
}