C++ 为什么没有 int128_t？
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Why isn't there int128_t?
提问by
A number of compilers provide 128-bit integer types, but none of the ones I've used provide the typedefs int128_t. Why?
许多编译器提供 128 位整数类型，但我使用过的编译器都没有提供 typedefs int128_t。为什么？
As far as I recall, the standard
据我所知，标准
- Reserves
int128_tfor this purpose - Encourages implementations that provide such a type to provide the typedef
- Mandates that such implementations provide an
intmax_tof at least 128 bits
int128_t为此目的准备金- 鼓励提供此类类型的实现提供 typedef
- 要求此类实现提供
intmax_t至少 128 位的
(and, I do not believe I've used an implementation that actually conforms to that last point)
（而且，我不相信我使用的实现实际上符合最后一点）
采纳答案by Keith Thompson
I'll refer to the C standard; I think the C++ standard inherits the rules for <stdint.h>/ <cstdint>from C.
我会参考C标准；我认为 C++ 标准从 C继承了<stdint.h>/的规则<cstdint>。
I know that gcc implements 128-bit signed and unsigned integers, with the names __int128and unsigned __int128(__int128is an implementation-defined keyword) on some platforms.
我知道 gcc在某些平台上实现了 128 位有符号和无符号整数，名称__int128和unsigned __int128(__int128是实现定义的关键字)。
Even for an implementation that provides a standard 128-bit type, the standard does not requireint128_tor uint128_tto be defined. Quoting section 7.20.1.1 of the N1570draft of the C standard:
即使对于提供标准 128 位类型的实现，该标准也不需要int128_t或被uint128_t定义。引用C标准N1570草案第7.20.1.1节：
These types are optional. However, if an implementation provides integer types with widths of 8, 16, 32, or 64 bits, no padding bits, and (for the signed types) that have a two's complement representation, it shall define the corresponding typedef names.
这些类型是可选的。但是，如果实现提供宽度为 8、16、32 或 64 位、无填充位且（对于有符号类型）具有二进制补码表示的整数类型，则应定义相应的 typedef 名称。
C permits implementations to defined extended integer typeswhose names are implementation-defined keywords. gcc's __int128and unsigned __int128are very similar to extended integer types as defined by the standard -- but gcc doesn't treat them that way. Instead, it treats them as a language extension.
C 允许实现定义的扩展整数类型，其名称是实现定义的关键字。gcc__int128和gcc与unsigned __int128标准定义的扩展整数类型非常相似——但 gcc 不会那样对待它们。相反，它将它们视为语言扩展。
In particular, if __int128and unsigned __int128were extended integer types, then gcc would be required to define intmax_tand uintmax_tas those types (or as some types at least 128 bits wide). It does not do so; instead, intmax_tand uintmax_tare only 64 bits.
特别是，如果__int128和unsigned __int128是扩展整数类型，则需要 gcc 将intmax_t和定义uintmax_t为这些类型（或定义为至少 128 位宽的某些类型）。它没有这样做；相反，intmax_t并且uintmax_t只有 64 位。
This is, in my opinion, unfortunate, but I don't believe it makes gcc non-conforming. No portable program can depend on the existence of __int128, or on any integer type wider than 64 bits.
在我看来，这是不幸的，但我不认为它会使 gcc 不符合标准。任何可移植程序都不能依赖于 的存在，也不能依赖于__int128任何大于 64 位的整数类型。
