c - Segmentation fault - char pointer

Question

In the code below, the line:

*end = *front;

gives a segmentation fault. I asked a similar question here but I'm not sure if this is because I have two copies of num. Please explain why it's seg-faulting. Thank you.

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

char* getPalin(char* num);

int main()
{
    char* num = (char*)malloc(100);

    num = "123456";

    printf("%s
", getPalin(num) );

    return 0;
}

char* getPalin(char* num)
{
    int length = strlen(num);

    if ( length % 2 == 0 )
    {
        char* front = num;
        char* end = num + strlen(num) - 1;  //pointer to end

        while( front != num + (length/2) )  //pointers not middle yet
        {
            *end = *front;

            printf("%c", *end);

            front++;
            end--;
        }
    }

    return num;
}

Answer 1

These two lines:

char* num = (char*)malloc(100);
num = "123456";

have the following effect.

The first allocates 100 bytes and sets num to point to those bytes.

The second changes num to point to the string "123456" which is almost certainly in read-only memory.

Any attempt to change the contents of read-only memory will result in a segmentation violation. You need to copy the string into the malloc'd num before attempting to change it, with:

strcpy (num, "123456");

That's the line you should have where you currently have:

num = "123456";