16 bytes-long integer types

Question

I'm compiling a C++ program using GCC on 64bits - machine/OS/ (with -m64 option passed to g++). As expected, sizeof(long double) == 16 – I'm wondering whether there is 16 bytes-long standard type for integers?

P.S. __int128_t is an artificial extension that emulates standard type as I understood. Other than that I didn't find anything.

`long double` isn't actually a 16-byte type. It probably uses 10 bytes to store the value, and the remainder is padding to make it 8-byte aligned. — Kerrek SB, Sep 18 '12 at 23:10
@Kerrek Is there a way to find out an exact size of the type then, i.e. how big can be the number that I can store in this type? — user1681646, Sep 18 '12 at 23:22
The `` traits are probably your best way to make portable statements about available ranges. — Kerrek SB, Sep 18 '12 at 23:24
[What does the C++ standard state the size of int, long type to be?](http://stackoverflow.com/q/589575/995714) — phuclv, May 29 '16 at 01:38
[`long double`](https://en.wikipedia.org/wiki/Long_double) can be 12 or 16 bytes in x86 gcc depending on which of [`-m96bit-long-double` and `-m128bit-long-double`](https://gcc.gnu.org/onlinedocs/gcc-3.3.1/gcc/i386-and-x86-64-Options.html) is selected, but the real content is only 80 bits, the remaining are just padding. On MSVC and many compilers it's exactly the same as 64-bit double. It may also be double-double, IBM extended double or IEEE-754 quadruple precision https://gcc.gnu.org/wiki/Ieee128PowerPC. No guarantee about it being 16 bytes either — phuclv, May 29 '16 at 01:45

score 3 · Answer 1 · answered Sep 18 '12 at 23:02

3

Nope, only guaranteed sizes are for char, unsigned char & signed char, and they are 1:

5.3.3 Sizeof [expr.sizeof]

1 The sizeof operator yields the number of bytes in the object representation of its operand. The operand is either an expression, which is not evaluated, or a parenthesized type-id. The sizeof operator shall not be applied to an expression that has function or incomplete type, or to an enumeration type before all its enumerators have been declared, or to the parenthesized name of such types, or to an lvalue that designates a bit-field. sizeof(char), sizeof(signed char) and sizeof(unsigned char) are 1. The result of sizeof applied to any other fundamental type (3.9.1) is implementation-defined. [ Note: in particular, sizeof(bool) and sizeof(wchar_t) are implementation-defined.74) —end note ] [ Note: See 1.7 for the definition of byte and 3.9 for the definition of object representation. —end note ]

(emphasis mine)

answered Sep 18 '12 at 23:02

Luchian Grigore

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I thought `long long` was guaranteed to have 64-bits? – Rapptz Sep 18 '12 at 23:09
@Rapptz not according to the standard. – Luchian Grigore Sep 18 '12 at 23:10
I thought so too (about long long). – user1681646 Sep 18 '12 at 23:11
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@Rapptz It's guaranteed to have _at least_ 64 bits. – Daniel Fischer Sep 18 '12 at 23:14
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To meet the minimum requirements of the C standard, `long long` must be _at least_ 64 bits long; it does not have to be exactly 64 bits. – Jonathan Leffler Sep 18 '12 at 23:17

score 1 · Answer 2 · answered May 29 '16 at 00:20

The c++ standard does not standardize the size of each type (accept char) but the difference in size in relation to each other.

A char has to be 1 byte.
A short has to be >= to a char.
An int has to be >= to a short.
A long has to be >= to an int.
and a long long has to be >= to a long.

same goes for float types.

I believe most compilers now support up to 8-byte integers

however I know for a fact some, if not most don't have 16-byte integers.

I also think that there are some object classes/structs that people have developed to emulate a 16-byte integer that would work on most compilers.

16 bytes-long integer types

2 Answers2