Thread: Setting character in malloc'ed char array

I am attempting to change a character in a character array.

In the code below, there are three attempts to do this. Only the first one will succeed. The last two both segfaults. If I understand correctly, str_one is declared in the heap, and could therefore be manipulated; and in contrast, str_two is declared in the stack and is therefore immutable, thus the segfault, when update it is attempted. However, I understand that using malloc, one is able to assign a pointer and allocate space in heap memory. Thus, I should be able to manipulate the assigned variable str_three. Doing so, however, results in a segfault.

I hope someone can shed light on this. Thanks!

Code:

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

int main (int argc, char const* argv[])
{
  char str_one[4092] = "This is string number one";
  char * str_two = "This is string number two";

  char * str_three;
  str_three = (char *) malloc(4092);
  str_three = "This is the third string";

  str_one[5] = 'X';
  printf("%s\n", str_one);

  str_two++;
  *str_two = 'X'; // segfaults here
  str_two--;
  printf("%s\n", str_two);

  str_three++;
  *str_three = 'X'; // segfaults here
  str_three--;
  printf("%s\n", str_three);


  return 0;
}

Originally Posted by SirPrattlepod

... String literals are constants, so when you try
and write to them you will/should get a segfault


Edit: to set str_three to contain a string (after you allocated memory for it), use the strcpy function

That's it! Thanks!