保证`long`与`size_t`一样宽 [英] Was `long` guaranteed to be as wide as `size_t`

问题描述

在寻找 unsigned long 足以容纳 size_t 的证据以作为对 printf 的论据的证据时，我遇到了两个事实(oids.

When looking for evidence of unsigned long being enough to hold size_t for the purpose of being argument to printf I ran into two fact(oid)s.

首先，有一个答案指出， long 的确不能保证足够大 size_t .另一方面，我看到了这个答案，建议使用 printf(％lu"，(无符号长)x)在C99之前的版本中， x 的大小为 size_t .

First there's this answer stating that long is indeed not guaranteed to be large enough for size_t. On the other hand I saw this answer suggesting to use printf("%lu", (unsigned long)x) in pre C99, x being of size_t.

所以问题是您可以假设还是可以保证 long 足以在 C99之前的版本中容纳 size_t .另一个问题是，是否可以保证 size_t 可以适用于任何其他标准化整数类型(除了明显的例外，例如 ssize_t ， ptrdiff_t 等).

So the question is could you assume or were you guaranteed that long were enough to hold size_t in pre C99. The other question is whether there exists any guarantee that size_t would fit in any of the other standardized integer types (except the obvious exceptions like ssize_t, ptrdiff_t and such).

推荐答案

没有这样的保证.

虽然通常对于 long 和 size_t 具有相同的大小是实现的，但并非总是如此.如注释中所述，Windows 64位的 long 和 size_t 具有不同的大小.

While it is common for implementation to have same size for long and size_t, it is not always the case. As put in the comments Windows 64-bit have different size for long and size_t.

还要注意，实现的 SIZE_MAX 的最小值是 65535 ，而 ULONG_MAX 的最小值是 4294967295 ( LONG_MAX 的 2147483647 ).(请注意， SIZE_MAX 随C99一起出现.)这意味着 size_t 可以保证至少为16位，但 unsigned long /长保证至少为32位.

Also notice that the minimum value of SIZE_MAX for an implementation is 65535 while the minimum value of ULONG_MAX is 4294967295 (2147483647 for LONG_MAX). (Note that SIZE_MAX appeared with C99.) It means that size_t is guaranteed to be at least 16-bit but unsigned long / long are guaranteed to be at least 32-bit.

在此答案之后，问题已经改变了一点……所以:

Question has changed a little bit after this answer... So:

所以问题是您是否可以假设还是可以保证足够长的时间来容纳C99之前的size_t.

即使在C89中也没有这样的保证. long 可以是32位，而 size_t 可以是64位.(请参见上面Windows 64位中带有MSVC的C89示例.)

There is no such guarantee even in C89. long can be 32-bit and size_t 64-bit. (See C89 example with MSVC in Windows 64-bit above.)

另一个问题是，是否可以保证size_t可以适合任何其他标准化整数类型(明显的例外，例如ssize_t，ptrdiff_t等).

同样，标准不提供此类保证. size_t 是另一个标准无符号整数类型的别名(并且它不能是扩展整数类型，因为C89没有扩展整数类型).

Again there is no such guarantee by the Standard. size_t is an alias for another standard unsigned integer type (and it cannot be an extended integer type as C89 does not have extended integer types).

这篇关于保证`long`与`size_t`一样宽的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！