strcat(dest, "\something") - backslash not showing

Question

I got a problem with C programming string concatenation.

Why strcat(dest, "\something") will not have the backslash copied to dest?

You may wish to follow the example.

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

int main()
{
    char *dest = "";
    char *test = "how \something";
    dest = (char *)malloc((strlen(test) + 1) * sizeof(char));

    strcat(dest, test);
    // Expect to see "how \something", but showed "how something"
    printf("%s\n", dest);

    free(dest);
    return 0;
}

Answer 1

backslash not showing

Escape it:

   char *test = "how \\something";

---

The compiler should have warned you about this like this (GCC 4.9.2):

main.c:8:16: warning: unknown escape sequence: '\s'
   char *test = "how \something";

---

Also this is not related to strcat().

To prove this just do

puts(test);

and receive the same output.

---

  but showed "how something"

Are you sure? I'd expected

how omething

---

Also

• sizeof (char) is 1 by definition.
• In C there is no need to cast void-pointers as for example returned by malloc(), nor is it recommended in any way.

So just do

  dest = malloc(strlen(test) + 1);

or so make the code safe against late changes of dest's type do

  dest = malloc((strlen(test) + 1) * sizeof *dest);

---

Addition

As pointed out by anatolyg in this comment you are concatenating to an uninitialised char-array here:

  strcat(dest, test);

This invokes undefined behaviour, which is not good.

To fix this either do

• dest[0] = '\0'; or
• use calloc() instead of malloc()

to 0-terminate dest and with this make the char-array a C-"string" before using it alike when passing it to strcat().

• Or just use strcpy() instead of strcat().