drop the LSB from an address in c

Question

I'm trying to drop (meaning zero the bit) the LSB from an address. The below piece of code which I wrote doesnt seem to do the thing I intended to do.

#include<stdio.h>
#include<stdlib.h>
#include<stdint.h>
struct node
{
    unsigned long key;
    struct node* child[2];
};
int main()
{
    struct node* node = (struct node*) malloc(sizeof(struct node));
    printf("%x\n",node);
    node = (struct node*)((uintptr_t) node >> 1UL);
    printf("%x\n",node);
}

You need to include in your question the output of the above code, why you feel that it is incorrect, and what the output you would like for it to be. — Bill Lynch, Jul 30 '14 at 17:08
It's not clear from your question if your intent is to zero the LSB or shift by 1, either of which could be called "dropping". Please clarify your question. — amo, Jul 30 '14 at 17:15
possible duplicate of [How do you set, clear and toggle a single bit in C/C++?](http://stackoverflow.com/questions/47981/how-do-you-set-clear-and-toggle-a-single-bit-in-c-c) — amo, Jul 30 '14 at 17:26

score 3 · Answer 1 · answered Jul 30 '14 at 17:05

3

If you want to clear the least significant bit, you can use:

(struct node*)((uintptr_t) node & (UINTPTR_MAX << 1))

but don't assign the result to node or you'll have a memory leak.

answered Jul 30 '14 at 17:05

ouah

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I'm already doing this as `(struct node*)((uintptr_t) p & FFFFFFFFFFFFFFFE)` – arunmoezhi Jul 30 '14 at 17:07
I wanted to know if I can do this without using the `&` operator and by just using the bit-wise operator – arunmoezhi Jul 30 '14 at 17:08
1

`&` is a bitwise operator, what do you mean? – ouah Jul 30 '14 at 17:09
:) I meant using the shift operator!! – arunmoezhi Jul 30 '14 at 17:09
You could use the shift operator twice: ((x >> 1) << 1). Not that that is a good solution. – Be Kind To New Users Jul 30 '14 at 17:12
@arunmoezhi see answer from Jashaszun but it will take one operation more – ouah Jul 30 '14 at 17:13
Ya got it thanks. I'll stick to `&` :) – arunmoezhi Jul 30 '14 at 17:13
3

@arunmoezhi but you should explain what do you want to achieve with that because it looks highly suspicious. – ouah Jul 30 '14 at 17:14
I make sure the address returned by `malloc` is byte-aligned and the lsb is always zero. So I use that as a flag in my code. So to get back the original address I just drop the bit. – arunmoezhi Jul 30 '14 at 17:16
@arunmoezhi The object returned by `malloc` is already guaranteed to be aligned for any use. – ouah Jul 30 '14 at 17:20
The flag part is explained in this post http://stackoverflow.com/questions/19389243/stealing-bits-from-a-pointer – arunmoezhi Jul 30 '14 at 17:22

score 1 · Answer 2 · answered Jul 30 '14 at 17:10

1

If you want to use the shift operator (as indicated in your comment on ouah's answer), you can use:

(struct node*)(((uintptr_t)node >> 1) << 1)

answered Jul 30 '14 at 17:10

Jashaszun

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score 1 · Answer 3 · edited Jul 30 '14 at 17:24

1

If your intent is to zero the low order bit of a value you would use an expression like:

x = z & ~1; // assuming x, y, and 1 are of the same bit width

This complements 1, turning it into something like 0b111111...1110 and performs a bitwise AND with the value z. The effect is the lowest bit is set to 0 while all others retain their values.

The expression you used, node >> 1UL, performs a shift -- that is, the bit in position 31 is moved to position 30, 30 to 29, ..., and 1 to 0, and the lowest order bit is discarded. It is usually equivalent to an integer division by 2. If this is what you intended, you should have gotten it.

edited Jul 30 '14 at 17:24

Jashaszun

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answered Jul 30 '14 at 17:11

user3889911

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Note that when you say `the bit in position 31 is moved to position 30`, that may not be the first bit moved (`1UL` does not need to be 32 bits). – Jashaszun Jul 30 '14 at 17:24
1

Also note that `~1` is equivalent to `-2`, and that both should be casted to the correct size. – Jashaszun Jul 30 '14 at 17:26

drop the LSB from an address in c

3 Answers3