Initialise array of character pointers inside a function to be used in main

Question

How can I initialise array of character pointers inside a function. It should be noted that length comes dynamically from other calculations inside the function. I cannot find length inside main(). For simplicity I have taken rand() to explain scenario.

Is there any way where I can initialise array of character pointers with certain length without passing length from main() to setArgs() ?

With below code I get errors:

temp.c:8:6: error: invalid use of array with unspecified bounds 8 | args[length];

#include<stdio.h>
#include<stdlib.h>

void setArgs(char* (*args)[])
{
    int length = (rand() % 5) + 2;

    args[length];
    args[0]="hello\0";
    args[1]="jack\0";
}
int main()
{
    char* myArgs[]={NULL};
        
    setArgs(&myArgs);

    for(int i=0;i<2;i++)
    {
        printf("%s", myArgs[i]);
    }

    return 0;
}

Answer 1

You have two options. Allocate the memory dynamically or first generate the length:

1.

#include<stdio.h>
#include<stdlib.h>

void setArgs(char *** args)
{
    int length = (rand() % 5) + 2;
    *args = malloc(sizeof(char *)*length);

    (*args)[0]="hello\0";
    (*args)[1]="jack\0";
}
int main()
{
    char ** myArgs = NULL;       
    setArgs(&myArgs);

    for(int i=0;i<2;i++)
    {
        printf("%s", myArgs[i]);
    }
    free(myArgs);
    return 0;
}

or 2.

#include<stdio.h>
#include<stdlib.h>

void setArgs(char ** args)
{
    args[0]="hello\0";
    args[1]="jack\0";
}
int main()
{
    int length = (rand() % 5) + 2;
    char* myArgs[length];
        
    setArgs(myArgs);

    for(int i=0;i<2;i++)
    {
        printf("%s", myArgs[i]);
    }
    return 0;
}

args[length]; doesnt change the length but just does an array-access at the index length