pointer

[cat5ive]

Hi,

char *string = "A string!";

why is *string+8 has a value of '!'?
Why not one position after '!' (which is a null char.)?
Since *string contain value of A(index 0), doesn't *string+8 mean add 8 more index to it?

Thanks

 

[Salem]

> why is *string+8 has a value of '!'?
Because your string has 9 characters, and ! is the last one


--

 

[cat5ive]

????please explain. From what I see, we start counting from 0 (which is 'A'), then '!' would be at 8 not 9. Or may be I should ask, what's the different between *string+8 and string[9]? (string[9] contain NULL)

 

[cat5ive]

got it now. It's the counting that get me (very)confuse. Sometime I'm not sure if I sould count from 0 or 1.

Thanks Salem.

 

[jstreich]

Indexing is zero based, size is simply howmany... So operations logically that would yeild a size start counting with 1, things that logically yeild a index start counting at 0.

Index:

012345678 9
A string![blue]NULL[/blue]
[code]
size = 9.

So...
char *string = "A string!"
string gets the address of the value A string.
string[0] gets the thing that is at the value of string (the address) + [0 (the offset in square braves) * size of datatype the pointer is pointing].
*(string+8)
8 plus that address is '!'.

 

[cat5ive]

Thanks a lot, jstreich.