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:
longandintremain 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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