C char pointer vs pointer to char array, increment dynamic allocation

Question

I am new to C, and I am having difficulty understanding the reason why the block of code below is not working.

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

int main()
{
    char *src = "http://localhost";

    /* THIS WORKS
       char scheme[10];
       char *dp = scheme;
     */

    //DOESN'T WORK
    char *dp = malloc(10);

    while (*src != ':') {
        *dp = *src;
        src++;
        dp++;
    }
    *dp = '\0';

    /* WORKS
       puts(scheme)
     */

    //DOESN'T WORK
    puts(dp);
}

The expected output is http to stdout. In both cases dp should be a pointer to an array of char pointers (char **). However it is printing nothing when using malloc method. I ran the code through GDB and my src and dp are getting erased 1 character at a time. If I enclose the while loop into a function call it works. I figured that the reason was because parameters are evaluated as a copy. However, then I read that arrays are the exception and passed as pointers. Now I am confused. I can work around this, but I am trying to understand why this way doesn't work.

Answer 1

You are changing dp inside the loop

dp = malloc(10);

let's say dp has the value 0x42000000

while () {
    dp++;
}

let's say the loop went 4 times, so dp has the value 0x42000004

*dp = 0;

now you put a null character at the address pointed to by dp

puts(dp);

and then you try to print that null :)

Save dp and print the saved value

dp = malloc(10);
saveddp = dp;
/* ... */
puts(saveddp);
free(saveddp); /* for completeness */

---

It works with scheme because scheme is an array and you cannot change that address!

Answer 2

After the end of the loop, dp points to the end of the allocated string. You need to save dp right after the malloc and increment the copy, not the original pointer to the beginning.