Question

What is the reason that gcc adds char* (e.g. “STRING”) and char (e.g. ‘C’) as pointers?

  const char *ccc = "Test1";
  const char t = 'T';
  const char *res = ccc + t;
  printf("%s, %p, %d, %p\n", res, ccc, t, res);

outputs

  , 0x8048d97, 84, 0x8048deb

I mean, can you point to the documentation, standard specs, or an article? Can I control or disable this behavior?

UPD: Why I ask and what is unexpected, is that

  CString() + 'c'

works as

 (char*)CString() + (char)char_var

when compiler cannot find appropriate operator +. I thought maybe to disable automatic concatenation and find all such places (in legacy code). But mostly I just wanted to find exact documentation for the behavior.

Answer 1

In ccc + t, t is treated as an integer. The net effect is that res points to ccc plus 84 bytes, where 84 is the ASCII code of 'T'.

It is worth pointing out that ccc + t operates purely on the pointers, and does not touch the actual string. I am saying this in case there’s any expectation that "Test" + 'T' might append the character to the string — it does not.