Declaring and copying an array of char strings in c - c

Question

I made a c program that attempts to add the values of one string array to another using a separate method:
#include <stdio.h>
#include <stdlib.h>
void charConv(char *example[])
{
example= (char* )malloc(sizeof(char[4])*6);
char *y[] = {"cat", "dog", "ate", "RIP", "CSS", "sun"};
printf("flag\n");
int i;
i=0;
for(i=0; i<6; i++){
strcpy(example[i], y[i]);
}
}
int main() {
char *x[6];
charConv( *x[6]);
printf("%s\n", x[0]);
}
However it keeps returning a segmentation fault. I'm only beginning to learn how to use malloc and c in general and its been puzzeling me to find a solution.

Answer 1

To pin-point your problem: you send *x[6] (here - charConv( *x[6]);) which is the first char of the 7'th (!!!) string (Remember, C is Zero-Base-Indexed) inside an array of 6 string you didn't malloc -> using memory you don't own -> UB.
Another thing I should note is char[] vs char * []. Using the former, you can strcpy into it strings. It would look like this:
'c' | 'a' | 't' | '\0' | 'd' | 'o' | 'g' | ... [each cell here is a `char`]
The latter ( what you used ) is not a contiguous block of chars but a array of char *, hence what you should have done is to allocate memory for each pointer inside your array and copy into it. That would look like:
0x100 | 0x200 | 0x300... [each cell is address you should malloc and then you would copy string into]
But, you also have several problems in your code. Below is a fixed version with explanations:
#include <stdio.h>
#include <stdlib.h>
void charConv(char *example[])
{
// example= (char* )malloc(sizeof(char[4])*6); // remove this! you don't want to reallocate example! When entering this function, example points to address A but after this, it will point to address B (could be NULL), thus, accessing it from main (the function caller) would be UB ( bad )
for (int i = 0; i < 6; i++)
{
example[i] = malloc(4); // instead, malloc each string inside the array of string. This can be done from main, or from here, whatever you prefer
}
char *y[] = {"cat", "dog", "ate", "RIP", "CSS", "sun"};
printf("flag\n");
/* remove this - move it inside the for loop
int i;
i=0;
*/
for(int i=0; i<6; i++){
printf("%s\t", y[i]); // simple debug check - remove it
strcpy(example[i], y[i]);
printf("%s\n", example[i]); // simple debug check - remove it
}
}
int main() {
char *x[6];
charConv( x); // pass x, not *x[6] !!
for (int i = 0; i < 6; i++)
{
printf("%s\n", x[i]); // simple debug check - remove it
}
}
As #MichaelWalz mentioned, using hard-coded values is not a good practice. I left them here since it's a small snippet and I think they are obvious. Still, try to avoid them

Answer 2

You need to start by understanding the pointers and some other topics as well like how to pass an array of strings to a function in C etc.
In your program, you are passing *x[6] in charConv() which is a character.
Made corrections in your program -
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void charConv(char *example[], int num)
{
int i;
for (i = 0; i < num; i++){
example[i] = (char* )malloc(sizeof(char)*4);
}
const char *y[] = {"cat", "dog", "ate", "RIP", "CSS", "sun"};
for(i = 0; i < num; i++){
strcpy(example[i], y[i]);
}
}
int main() {
char *x[6];
int i = 0;
charConv(x, 6);
/* Print the values of string array x*/
for(i = 0; i < 6; i++){
printf("%s\n", x[i]);
}
return 0;
}