This malloc shouldn't work

Question

Here is my code.

 int     main()
  {
  char *s;
  int i = 0;

  printf(\"%lu \\n\", sizeof(s));

  s = malloc(sizeof(char) * 2);

  printf(\"%lu \\n\", sizeof(s         


        

Answer 1

In illustration (and complete agreement with) @Ed S's answer, Try this example of your code with the small additions of an additional variable declared exactly the same way and malloc'ed right after the char *s.

Although no guarantees that the variables are stored sequentially in memory, creating them this way makes it a high probability. If so, char *t will now own the space that char *s will encroach on, and you will get a seg fault:

int     main()
{
    char *s;
    char *t;//addition
    int i = 0;

    printf("%lu \n", sizeof(s));

    s = malloc(sizeof(char) * 2);
    t = malloc(sizeof(char) * 2);//addition

    printf("%lu \n", sizeof(s));
    /*Why is this working?*/
    while (i <= 5)
    {
      s[i] = 'l';
      i++;
    }
    printf("%s \n", s);

    printf("%lu \n", sizeof(char));

    printf("%lu \n", sizeof(s[0]));
}

Note: In the environment I use, (Windows 7, NI Run-Time, debug, etc) I get a seg-fault either way, which somewhat supports the undefined behavior assertions in other answers.