Is there a way to check if char * is null with bitwise

Question

I have a char array. Some of the elements allocated, and some are not. I want to count all those not allocated without asking if (!arr[i])

char* arr[500];
.
.
.
for(i=0; i < 500; ++i)
{
   if( !arr[i] )
   { 
      counter++;
   }
}

but making some bitwise magic like this in C

for(i=0; i < 500; ++i)
{
    counter += !(arr[i]| 0x00); //this does not work 
}

Maybe the question is more general: is there a way to compare any pointer with NULL with bitwise operations.

Edit: no, not with ==; I understand how to do it with ==, but I assume when it is compiled, it does not differ from if(!arr[i]) in assembler.

What "bitwise"? Why do you even... I mean, it doesn't make sense. — , Dec 30 '13 at 13:48
@fasked I assume == translated into something similar to if(!arr[i]) — Dabo, Dec 30 '13 at 13:48
possible duplicate of [How to calculate the non-zero elements in the array?](http://stackoverflow.com/questions/20839793/how-to-calculate-the-non-zero-elements-in-the-array) — , Dec 30 '13 at 13:49
(And nobody can be bothered to do their own friggin' homework.) — , Dec 30 '13 at 13:49
@DavidBo It's not possible, because `NULL` is not has to be represented with all zero-bits. — fasked, Dec 30 '13 at 14:16
Step back, think _gloves_... isn't `if (!arr[i])` easy to read? why would you want to make life more complicated than it already is? — Elias Van Ootegem, Dec 30 '13 at 14:43
wanted to know if it is possible after reading this http://stackoverflow.com/a/11227902/2549281 — Dabo, Dec 30 '13 at 14:50
@fasked: the null pointer *constant* is always 0-valued, hence the `NULL` macro is always going to be zero-valued (6.3.2.3/3 plus footnote 66); the corresponding null pointer *value* does not have to be zero-valued, but in the context of the source code, you're dealing with the null pointer *constant*. Having said that, `!arr[i]` or `arr[i] == NULL` are both safer than using a bitwise operation. — John Bode, Dec 30 '13 at 14:58
@JohnBode Yes, that is what I meant. `0` integer *value* and `NULL` pointer *value* may be not equal bit by bit. — fasked, Dec 30 '13 at 15:19
@DavidBo In your "this does not work loop", you might need to explain how it doesn't work. Since `x | 0 == x`, your `arr[i] | 0x00 == arr[i]`, so the code is no different than `counter += !(arr[i])`... — twalberg, Dec 30 '13 at 15:30
@twalberg compiler does not allow to make bitwise OR between pointer and 0. it gives error : invalid operands to binary | (have ‘char *’ and ‘int’). So i wondered if there still some syntax that allows it. — Dabo, Dec 30 '13 at 15:38
@DavidBo Ah, right... There is specific syntax to force it to work (hint: casts), but, as I mentioned, because `x | 0 == x` always, there's not really any point in it. The compiler would likely generate nearly identical code to your working example, even if you did force it - it would likely know that there's no point to OR with a constant 0, and so would optimize it away... — twalberg, Dec 30 '13 at 15:46
@twalberg it actually did work, thanks for your hint. for(i=0; i < 500; i++) { counptr += !( (int)arr[i] | 0x0 ); printf("%d\n",counptr); } — Dabo, Dec 31 '13 at 08:48
Dabo, please explain how reading the branch prediction question made you think that bitwise operations are superior? — Leeor, Feb 28 '14 at 13:46
@Leeor not superior, but in the answer to the linked question presented bitwise technique reduced execution time, because `if` was eliminated from the loop. — Dabo, Feb 28 '14 at 15:14

score 3 · Answer 1 · edited Dec 30 '13 at 14:07

3

for(i=0; i < 500; ++i)
{
    counter += arr[i] == NULL;
}

Or even...

for(i=0; i < 500; ++i)
{
    counter += !arr[i];
}

edited Dec 30 '13 at 14:07

answered Dec 30 '13 at 13:47

mcleod_ideafix

11,128
2
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I suspect that casting to `bool` is not portable to systems with non-zero NULL pointers. – user694733 Dec 30 '13 at 13:57
1

@user694733 However, `!arr[i]` is always true if `arr[i]` is a `NULL` pointer, regardless to the actual numeric value of the null pointer. – Dec 30 '13 at 14:07

Is there a way to check if char * is null with bitwise

1 Answers1