How can I simplify these char array? [closed]

Question

Hello I Have to loop this char array but the last argument changes by one everytime. The last number is supposed to be 100 and since i cant write 100 times the same line of argument i wanted to see if there is a simpler way of writing it

#include <iostream>
using namespace std;
int main(int argc, char* argv[])

char command[100]= "ousbsim -r io PORTB 1";
char command2[100]= "ousbsim -r io PORTB 2";
char command3[100]= "ousbsim -r io PORTB 3";
char command4[100]= "ousbsim -r io PORTB 4";
char command5[100]= "ousbsim -r io PORTB 5";
char command6[100]= "ousbsim -r io PORTB 6";

and so on..

Answer 1

Simply create an array of strings, and then use a loop to populate the strings.

For example, try something like this:

#include <cstdio>

int main()
{
    char commands[100][36];

    for (int i = 0; i < 100; ++i)
    {
        std::sprintf(commands[i], "ousbsim -r io PORTB %d", i+1);
    }

    // use commands[0]..commands[99] as needed...

    return 0;
}

Or, using C++'s std::string instead of C-style char[] arrays:

#include <string>

int main()
{
    std::string commands[100];

    for (int i = 0; i < 100; ++i)
    {
        commands[i] = "ousbsim -r io PORTB " + std::to_string(i+1);

        /* alternatively:
        #include <sstream>
        std::ostringstream oss;
        oss << "ousbsim -r io PORTB " << i+1;
        commands[i] = oss.str();
        */

        /* alternatively:
        #include <format>
        commands[i] = std::format("ousbsim -r io PORTB {}", i+1);
        */
    }

    // use commands[0]..commands[99] as needed...

    return 0;
}