I've seen this done in C before:
#define MY_STRING "12345"
...
#define SOMETHING (MY_STRING + 2)
What does SOMETHING get expanded to, here? Is this even legal? Or do they mean this?:
#define SOMETHING (MY_STRING[2])
views:
393answers:
2
+13
A:
When you have an array or pointer, p+x is equivalent to &p[x]. So MY_STRING + 2 is equivalent to &MY_STRING[2]: it yields the address of the third character in the string.
Notice what happens when you add 0. MY_STRING + 0 is the same as &MY_STRING[0], both of which are the same as writing simply MY_STRING since a string reference is nothing more than a pointer to the first character in the string. Happily, then, the identity operation "add 0" is a no-op. Consider this a sort of mental unit test we can use to check that our idea about what + means is correct.
John Kugelman
2010-08-25 22:22:43
I think my hangup here was that I didn't know that a quoted string was equal to the character array, at least syntax-wise. Thanks. +1.
Joe
2010-08-25 22:34:16
+33
A:
String literals exist in the fixed data segment of the program, so they appear to the compiler as a type of pointer.
+-+-+-+-+-+--+
|1|2|3|4|5|\0|
+-+-+-+-+-+--+
^ MY_STRING
^ MY_STRING + 2
Ignacio Vazquez-Abrams
2010-08-25 22:23:42
@Hamish Grubijan: Yes, the same will go for Unicode. The L"StringLiteral" will have `const wchar_t*` type, so `+` will advance by the number of `wchar_t` s.
sharptooth
2010-08-26 05:23:15
Throw some votes on the other answer as well, because it's a good complement.
Skurmedel
2010-08-26 08:46:46
Men are visual creatures; if it looks good, then it must be right.
Hamish Grubijan
2010-08-26 13:34:22