Bit field memory

Question

If i declare something like this

struct S{
  unsigned int bit:4;
}

How is it working?

I allocate 2 bytes in memory(size of structure(got this size from here http://en.cppreference.com/w/cpp/language/bit_field) but use only 4 bits of it, and other memory in that structure is wasted.
I allocate only 4 bits, nothing more.

I'm very confused about this and can't find any info about this anywhere.

Ok, thanks guys, I wasn't sure about that thing, thanks again for help :) — bingo157, Apr 25 '15 at 16:40
Fundamentally, you are asking for **at least 4 bits**, the compiler is allowed to allocate those 4 bits in a larger container, such as a 64-bit integer. As long as the compiler supplies you with 4 bits, the compiler conforms to the language. — Thomas Matthews, Apr 25 '15 at 18:36
Not exactly the first option. The compiler may allocate 2, 4, 8 or more bytes because of the `unsigned int`. — Thomas Matthews, Apr 25 '15 at 19:35

Beta Carotin · Answer 1 · 2015-04-25T16:52:20.233

3

When you write

S s;

you allocate sizeof (S) bytes, which seems to be 2 in your case.
The fact that you only use 4 bits of that space does not change the size.

edited Apr 25 '15 at 16:52

answered Apr 25 '15 at 16:35

Beta Carotin

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I removed the edit that was done on this post because this is C++, not C. No need to write `struct` in front of the type name. – Beta Carotin Apr 25 '15 at 16:53
Ah, didn't know that. On which modern architecture `sizeof(struct S);` will be `2`? – hek2mgl Apr 25 '15 at 16:53
@hek2mgl I don´t know why `sizeof (S)` would be `2`, but for the OT it seems to be the case :) – Beta Carotin Apr 25 '15 at 16:55
On `32` and `64` bit platforms `sizeof(S)` will return `4`. – hek2mgl Apr 25 '15 at 16:56
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On 16-bit processors, the `sizeof(int)` would be 2, assuming the `sizeof(char)` is one on those systems. – Thomas Matthews Apr 25 '15 at 18:31
Whatever guys, I've got this info about sizeof structure from that link which is in the question. – bingo157 Apr 25 '15 at 18:46

hek2mgl · Answer 2 · 2015-04-25T16:42:18.170

0

You cannot allocate 4 bits. The lowest memory allocation unit is a byte, 8 bits.

However there is no reason to use an unsigned int this would waste even more memory, use an unsigned char.

edited Apr 25 '15 at 16:42

answered Apr 25 '15 at 16:36

hek2mgl

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More like `CHAR_BIT` bits to be exact. – Beta Carotin Apr 25 '15 at 16:37

score 0 · Answer 3 · answered Apr 25 '15 at 16:38

0

This will allocate 4 bytes of memory and only use 4 bits of it. Bit fields are required to take up all the space for the type they are declared as, even if it isn't all used.

If you want to use the memory more efficiently you could do this:

struct S{
    unsigned char bit : 4;
};

and it would only allocate a byte.

Also sizeof(unsigned int) is usually 4 bytes so sizeof(S) will be at least 4 bytes.

answered Apr 25 '15 at 16:38

phantom

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Whether it allocates 8, 4, or 2 bytes for an integer is platform dependent. The size of an `int` is minimally 16 bits to satisfy the range requirements. Most compilers use the processor's word size for an `int`, which is 8 bytes on a 64-bit system, 4 bytes on a 32-bit system and 2-bytes on 16 and 8-bit systems. The is no maximum size limit. – Thomas Matthews Apr 25 '15 at 18:34

Bit field memory

3 Answers3