What is the bit size of long on 64-bit Windows?(在 64 位 Windows 上 long 的位大小是多少?)
问题描述
不久前,有人告诉我 long
在 64 位机器上不是 64 位,我应该始终使用 int
.这对我来说没有意义.我看过文档(例如 Apple 官方网站上的文档)说在为 64 位 CPU 编译时,long
确实是 64 位.我在 64 位 Windows 上查找了它,发现
Not to long ago, someone told me that long
are not 64 bits on 64 bit machines and I should always use int
. This did not make sense to me. I have seen docs (such as the one on Apple's official site) say that long
are indeed 64 bits when compiling for a 64-bit CPU. I looked up what it was on 64-bit Windows and found
- Windows:
long
和int
保持 32 位长度,以及特殊的新数据类型为 64 位整数定义.
- Windows:
long
andint
remain 32-bit in length, and special new data types are defined for 64-bit integers.
(来自 http://www.intel.com/cd/ids/developer/asmo-na/eng/197664.htm?page=2)
我应该使用什么?如果不在 Windows 上,我是否应该将 uw
、sw
((un)signed width) 定义为 long
,否则请检查目标 CPU 位大小?
What should I use? Should I define something like uw
, sw
((un)signed width) as a long
if not on Windows, and otherwise do a check on the target CPU bitsize?
推荐答案
在 Unix 世界中,对于 64 位平台,整数和指针的大小有几种可能的安排.最广泛使用的两个是 ILP64(实际上,只有极少数的例子;Cray 就是这样一个例子)和 LP64(几乎所有其他的例子).缩写词来自int, long, pointers are 64-bit"和long, pointers are 64-bit".
In the Unix world, there were a few possible arrangements for the sizes of integers and pointers for 64-bit platforms. The two mostly widely used were ILP64 (actually, only a very few examples of this; Cray was one such) and LP64 (for almost everything else). The acronynms come from 'int, long, pointers are 64-bit' and 'long, pointers are 64-bit'.
Type ILP64 LP64 LLP64
char 8 8 8
short 16 16 16
int 64 32 32
long 64 64 32
long long 64 64 64
pointer 64 64 64
ILP64 系统被 LP64 所取代(也就是说,几乎根据 Aspen 小组的建议,所有后来的参赛者都使用 LP64;只有具有 64 位操作传统的系统才使用不同的方案).所有现代 64 位 Unix 系统都使用 LP64.MacOS X 和 Linux 都是现代 64 位系统.
The ILP64 system was abandoned in favour of LP64 (that is, almost all later entrants used LP64, based on the recommendations of the Aspen group; only systems with a long heritage of 64-bit operation use a different scheme). All modern 64-bit Unix systems use LP64. MacOS X and Linux are both modern 64-bit systems.
Microsoft 使用不同的方案转换到 64 位:LLP64('long long,指针是 64 位').这意味着 32 位软件可以在不改变的情况下重新编译.它的缺点是与其他人所做的不同,并且还需要修改代码以利用 64 位容量.总是需要修改;它只是一组与 Unix 平台所需的不同的修订.
Microsoft uses a different scheme for transitioning to 64-bit: LLP64 ('long long, pointers are 64-bit'). This has the merit of meaning that 32-bit software can be recompiled without change. It has the demerit of being different from what everyone else does, and also requires code to be revised to exploit 64-bit capacities. There always was revision necessary; it was just a different set of revisions from the ones needed on Unix platforms.
如果您围绕平台中立的整数类型名称设计软件,可能使用 C99
If you design your software around platform-neutral integer type names, probably using the C99 <inttypes.h>
header, which, when the types are available on the platform, provides, in signed (listed) and unsigned (not listed; prefix with 'u'):
int8_t
- 8 位整数int16_t
- 16 位整数int32_t
- 32 位整数int64_t
- 64 位整数uintptr_t
- 大到足以容纳指针的无符号整数intmax_t
- 平台上最大整数大小(可能大于int64_t
)
int8_t
- 8-bit integersint16_t
- 16-bit integersint32_t
- 32-bit integersint64_t
- 64-bit integersuintptr_t
- unsigned integers big enough to hold pointersintmax_t
- biggest size of integer on the platform (might be larger thanint64_t
)
然后,您可以在重要的地方使用这些类型对您的应用程序进行编码,并对系统类型(可能不同)非常小心.有一个 intptr_t
类型——一个用于保存指针的有符号整数类型;您应该计划不使用它,或者仅将其用作两个 uintptr_t
值 (ptrdiff_t
) 相减的结果.
You can then code your application using these types where it matters, and being very careful with system types (which might be different). There is an intptr_t
type - a signed integer type for holding pointers; you should plan on not using it, or only using it as the result of a subtraction of two uintptr_t
values (ptrdiff_t
).
但是,正如问题所指出的那样(难以置信),64 位机器上整数数据类型的大小有不同的系统.习惯它;世界不会改变.
But, as the question points out (in disbelief), there are different systems for the sizes of the integer data types on 64-bit machines. Get used to it; the world isn't going to change.
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