How can I populate a new array with strings passed as command line arguments in argv[]?

I am relatively new to coding, especially in C. I am attempting to create a simple program designed to take a phrase, or list of strings, as command line arguments that populate char* argv[]. I then want to perform simple ASCII based changes to each character (such as making a => b, b => c, … z => a etc.), and then populate a new array with the altered strings. Essentially, the new object would be an array of arrays of characters, just like I believe argv[] to be.

Here is my code, although I have been bombarded with a variety of "pointer incompatible conversion assignment **char to string int char variable…..Segmentation fault" style errors.

#include <stdio.h>  
#include <string.h>  


int main(int argc, char* argv[])  
{  
    
    char* cipher[argc];  
    for (int i = 1; i < argc; i++)  
    {  
        int n = strlen(argv[i]);  
        for (int j = 0; j < n-1; j++)  
        {  
            cipher[i][j] = argv[i][j];  
        }  
    }  
    for (int i = 0; i<argc-1; i++)  
    {  
        printf("%s\n", cipher[i]);  
    }  
}  

The code would be run via command line as follows:

./secret the man in the middle is guilty

And then return a scrambled form of this. I have omitted the scrambling algorithm because this was not something I was having issues with.

I understand that I could more easily create code that prints the altered message, but I want to be able to store the new message in an array in case of further alterations. I’m not understanding why I can’t just access the values of characters of specific 2d indexes and populate the new "cipher" array with them.

What am I not understanding?

>Solution :

Bunch of issues:

#include <stdio.h>  
#include <string.h>  


int main(int argc, char* argv[])  
{  
    

    // first issue: this is only allocating space for "pointers" to
    // strings [argc many] but no space for actual strings so you have to 
    // allocate them.


    // also this is only really workable with c99 (assuming that's the case)
    char* cipher[argc];  

    // this is okay 
    for (int i = 1; i < argc; i++)  
    {  

       // this is where you would allocate space and 
       // perform the copy
        int n = strlen(argv[i]);  
        for (int j = 0; j < n-1; j++)  
        {  
            cipher[i][j] = argv[i][j];  
        }  
    }

   
    for (int i = 0; i<argc-1; i++)  
    {  
        printf("%s\n", cipher[i]);  
    }  
}  

So at least minimally:

#include <stdio.h>
#include <string.h> 

int main(int argc, char* argv[])  
{  
   
    char* cipher[argc];

    // this is okay 
    for (int i = 1; i < argc; i++)  
    {  
         // strdup will return a newly allocated copy
         // and since cipher[] is an array of pointers 
         // you are good to go.
         cipher[i-1] = strdup(argv[i]);
    }

   
    for (int i = 0; i<argc-1; i++)  
    {  
        printf("%s\n", cipher[i]);  
    }  

    // later on after you are finished with cipher 
    // assuming that you still have other things to do
    // you can de-allocate the memory (although not sure if
    // that will be required)

    for(int i = 0; i < argc - 1; i++) {
         free(cipher[i]); // strdup allocated memory cleanup
    }

    // you can no longer use cipher[i] here because it's contents
    // have been destroyed by the free above.
}