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!