Is sizeof(int) always equal to sizeof(void*)

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sizeof (int) == sizeof (void*)?

I was wondering whether it is guaranteed that, in both 32-bit and 64-bit systems, sizeof(int) is always equal to sizeof(void*) (i.e. 32 and 64 bits respectively).

Additionally, I need to know whether it is always guaranteed that a long int can accommodate the bits of an int and a void* together, e.g.

long int lint = (((int)integer)<<sizeof(int)) | (void*)ptr;

`long int` and `int` actually have the *same* size on many platforms. — Kerrek SB, Feb 11 '12 at 22:03
Perhaps if you told us what you were *really* trying to accomplish, we could answer with some more accurate solution. — André Caron, Feb 11 '12 at 22:09
@LightnessRacesinOrbit: "No" is definitely accurate, but it's not a solution. At least, not to whatever OP is trying to accomplish. — André Caron, Feb 11 '12 at 23:07
@AndréCaron: This is a programming Question & Answer site, not a Problem & Solution site :) — Lightness Races in Orbit, Feb 11 '12 at 23:08
@LightnessRacesinOrbit: I don't see the point in answering the question if it doesn't solve OP's problem. — André Caron, Feb 12 '12 at 00:02
@AndréCaron: The OP has not stated a problem. The OP has asked a question. The OP will gain knowledge from the answer to the question. Pretty simple stuff. — Lightness Races in Orbit, Feb 12 '12 at 00:21

score 7 · Answer 1 · answered Feb 11 '12 at 22:00

I was wondering whether it is guaranteed that, in both 32-bit and 64-bit systems, sizeof(int) is always equal to sizeof(void*)

No.

I need to know whether it is always guaranteed that a long int can accommodate the bits of an int and a void* together

No. A quick proof is to consider that sizeof(long int) == sizeof(int) on many modern platforms, probably including the one you're using.

The more important question is why you think you "need to know" this; the fact that you're asking such questions makes me concerned that your code is likely to be ... wobbly.

To give an example for the first "no": Using gcc `sizeof(int)` is 4 on 64-bit platforms (whereas `sizeof(void*)` is 8 of course). — sepp2k, Feb 11 '12 at 22:14

mah · Answer 2 · 2012-02-11T22:14:22.887

2

The size of an int is implementation dependent and though it may turn out to be equal to the size of a pointer in many systems, there is no guarantee.

If you decide you need code to depend on this, you can include something such as:

if (sizeof(int) != sizeof(void *))
{
    fprintf(stderr, "ERROR: size assumptions are invalid; this program cannot continue.\n");
    exit(-1);
}

edited Feb 11 '12 at 22:14

answered Feb 11 '12 at 22:03

mah

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Alas, no - the preprocessor doesn't know what something's size is. I'll edit; thanks. – mah Feb 11 '12 at 22:13
Personally I don't like assert() as I've seen it #if'd out of non-debug releases too often. – mah Feb 12 '12 at 12:16
why not `static_assert`? this assumption won't change at runtime. – Janus Troelsen Aug 10 '14 at 16:43

score 1 · Answer 3 · answered Feb 11 '12 at 22:20

1

Recent C99 standard provides a <stdint.h> header defining an intptr_t integral type guaranteed to have the same size as pointers.

On my Debian/AMD64/Sid, sizeof(int) is 4 bytes, but sizeof(void*) and sizeof(intptr_t) and sizeof(long) are all 8 bytes.

answered Feb 11 '12 at 22:20

Basile Starynkevitch

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score 1 · Accepted Answer · answered Feb 11 '12 at 22:24

I was wondering whether it is guaranteed that, in both 32-bit and 64-bit systems, sizeof(int) is always equal to sizeof(void*) (i.e. 32 and 64 bits respectively).

No.

Additionally, I need to know whether it is always guaranteed that a long int can accommodate the bits of an int and a void* together

No.

What you are looking for is: std::intptr_t

sizeof(std::intptr_t) == sizeof(void*)

std::intptr_t is defined as an integer of a size sufficient to hold a pointer.

Technically its optional part of the standard.
But you can usually find in the header file <cstdint> See: 18.4.1 Header <cstdint> synopsis [cstdint.syn]

Is sizeof(int) always equal to sizeof(void*)

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