Any predefined type for unsigned 96 bit integer?

Question

I am working on a feature which needs to keep track of a bit-map of length 96. I use this map to program asic below. Is there a standard integer type to hold 96 bits. For 64 bits, we have unsigned long long. Anything similar for 96 bits? Any alternate suggestion welcome as well.

PS: This is Cisco OS based on linux. Language is C.

i think the answer would be no ( because there is no existing definition), but how do i exactly look for it? — Deepanjan Mazumdar, Jul 13 '12 at 18:05

score 3 · Answer 1 · answered Jul 13 '12 at 17:30

3

GCC has __int128 for 128-bit integers on targets that support it, but nothing for 96 bits specifically.

answered Jul 13 '12 at 17:30

Ignacio Vazquez-Abrams

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+1, but to be a bit more precise, gcc has that on architectures that have a native 64 bit integer type. – Jens Gustedt Jul 13 '12 at 17:35

score 3 · Accepted Answer · answered Jul 13 '12 at 17:40

3

I'd probably go with an array of 3 uint's. That should be fast enough, and not a lot more complex.

EG, to set a bit:

wordNo = i / 32
bitNo = i - (32*wordNo)
mask = 2 ** bitNo

array[wordNo] |= mask

...or thereabout.

answered Jul 13 '12 at 17:40

user1277476

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qwertz · Answer 3 · 2012-07-15T10:22:33.847

There is a way to create one variable that size is exact 96 bits: Bitfields in a union or struct.

typedef union { // you can use union or struct here
    __uint128_t i : 96;
} __attribute__((packed)) uint96_t;

uint96_t var;  // new uint96_t variable
var.i = 123;   // set the value to 123 (for example)

This worked for me with gcc. If you test the size of uint96_t with sizeof you should get 12 bytes (12 * 8 = 96).

score 0 · Answer 4 · answered May 04 '20 at 00:25

0

You should use clang's new feature _ExtInt. It is in clang 11, so you may not have it. Look here for details and usage.

answered May 04 '20 at 00:25

xilpex

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Any predefined type for unsigned 96 bit integer?

4 Answers4

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