The #define variable usage

Question

I am working on a project written in C programming language. I got a code snippet as below

unsigned char value[10];
#define arr() (&value[0])

Why have they defined a "function"(arr()) kind of #define'd variable for an unsigned char array?

They are trying to use the variable like arr()[1],arr()[2] etc.,

Is arr() + 2 equal to value + 2. I tried executing a small program, but theses two results gave me different answers. how is it possible. Because they are assigning the address of first array to arr(). Should these two not be the same?

Can anyone explain what is the significance of defining a variable as above?

That's not a function, that's a [macro](http://gcc.gnu.org/onlinedocs/cpp/Macros.html). And can you show the small program you used that gave you different answers? They should give the same. — David Robinson, Mar 27 '13 at 05:13
I think it would take a psychic to know why they `#define`d `arr()` like that. — Cornstalks, Mar 27 '13 at 05:18
@Cornstalks its possible they wanted to decay the array to a pointer. However, that doesn't explain why they wanted a function-like macro — Richard J. Ross III, Mar 27 '13 at 05:22

score 1 · Answer 1 · answered Mar 27 '13 at 05:16

1

I can't tell you why they've done it, but yes, arr() + 2 and value + 2 are the same thing.

answered Mar 27 '13 at 05:16

caf

233,326
40
323
462

score 1 · Answer 2 · answered Mar 27 '13 at 05:18

This line #define arr() (&value[0]) means that whenever the preprocessor (which runs before the compiler) encounters arr() it replaces it in the file with (&value[0]).

So arr()[1] means (&value[0])[1]. value[0] is what's stored at the first index of value, & takes its address of it again... which is the same as value. So it should just be value[1] overall, unless I am missing something.

score 1 · Answer 3 · edited May 23 '17 at 12:04

1

arr()[i] ==> (&value[0])[i] ==> value[i]

Also

arr() + i ==> (&value[0]) + i ==> value + i

So

arr() + 2 ==> (&value[0]) + 2 ==> value + 2

_{I can only guess that programmer writing this way to make coding uniform with other part of hist code like this example}

edited May 23 '17 at 12:04

Community

1
1

answered Mar 27 '13 at 05:18

Grijesh Chauhan

57,103
20
141
208

Barath Ravikumar · Answer 4 · 2013-03-27T07:14:30.083

1

Is arr() + 2 equal to value + 2. I tried executing a small program, but these two results gave me different answers. how is it possible. Because they are assigning the address of first array to arr(). Should these two not be the same?

Yes they are the same.

Try this program

#include <stdio.h>

#define arr() (&value[0])

int main(void)
{

unsigned char value[11] = "0123456789";
printf("arr()[2] = %c\n",  arr()[2] );
printf("arr()+ 2 = %c\n",*(arr() + 2) );
printf("value+ 2 = %c\n",*(value + 2) );

return 0;
}

edited Mar 27 '13 at 07:14

answered Mar 27 '13 at 05:32

Barath Ravikumar

5,658
3
23
39

You should write `value[11]`, as the string `"0123456789"` has actual length `11` including `\0`. – Alex Mar 27 '13 at 07:11

The #define variable usage

4 Answers4