我可以信任的sizeof（为size_t）＆LT; = sizeof的（unsigned long int类型）始终是真的吗？ [英] Can I trust sizeof(size_t) <= sizeof(unsigned long int) is always true?

问题描述

我可以信任的sizeof（为size_t）＆LT; = sizeof的（unsigned long int类型）始终是真实的，按C89标准

Can I trust sizeof(size_t) <= sizeof(unsigned long int) is always true, according to C89 standard?

也就是说，我不会，如果我使用洛斯值unsigned long类型，其中为size_t 预计，反之亦然

i.e., I will not loos value if I use a unsigned long where size_t is expected and vice-versa.

推荐答案

也许与为size_t 整数类型发生了什么，并在 C89 是它的基本原理文档。阅读本文件的仔细的第3.4.4节：

Perhaps the best explanation of what happens with size_t and integer types in C89 is the rationale document of it. Read carefully the section 3.4.4 of this document:

的sizeof和为size_t为C89（理）

第三段说：

类型的sizeof，不管它是什么，发表的（在库
  头），为为size_t，因为它是为程序员有用
  能够参考此类型。这一要求隐
  限制为size_t是为现有的无符号整数的同义词
  型，从而撤销任何概念，最大的申报对象
  可能过大跨越，即使一个unsigned long。

The type of sizeof, whatever it is, is published (in the library header ) as size_t, since it is useful for the programmer to be able to refer to this type. This requirement implicitly restricts size_t to be a synonym for an existing unsigned integer type, thus quashing any notion that the largest declarable object might be too big to span even with an unsigned long.

这意味着，对于 C89 ，一般为size_t 是一样的preexisting 无符号整数型，C89什么意味着 unsigned char型之一，无符号短，无符号整型，无符号长。结果
特别是，类型的每个值为size_t 是范围无符号长。结果
通过阅读标准C89的规格，也可以是看到的sizeof（为size_t）＆LT; = sizeof的（长）。

This means that, for the concern of C89, in general size_t is the same as a preexisting unsigned integer type, what in C89 implies one of the unsigned char, unsigned short, unsigned int, unsigned long.
In particular, every value of type size_t is in the range of unsigned long.
By reading the specifications of the standard C89, you can see also that sizeof(size_t)<=sizeof(long).

现在，在这种情况 C99 有一点不同。该标准指出：

Now, the situation in C99 is a little different. This standard says that:

1. 7.17相提并论。 2 为size_t 是一个无符号整数类型。


2. 7.17相提并论。 4用于类型为size_t [...]不应该有一个整数转换等级大于签订长整型


3. ---------除非实现支持大到足以使这个必要的对象。



4. 6.2.5相提并论。 8对于任何两个整数类型具有相同的符号性和不同的整数转换等级（见6.3.1.1），以较小的整数转换等级的类型值的范围是其他类型的值的子范围。

1. 7.17 par. 2 size_t is an unsigned integer type.
2. 7.17 par. 4 The types used for size_t [...] should not have an integer conversion rank greater than that of signed long int
3. --------- unless the implementation supports objects large enough to make this necessary.

4. 6.2.5 par. 8 For any two integer types with the same signedness and different integer conversion rank (see 6.3.1.1), the range of values of the type with smaller integer conversion rank is a subrange of the values of the other type.

由于整数转换的等级签订长整型是相同的 unsigned long int类型，这意味着为size_t 的值的范围包含在 unsigned long int类型的值的范围。
但是，在上面的列表中的项目3敞开了大门，以此规则的例外。

Since the integer conversion rank of signed long int is the same that unsigned long int, this implies that the range of values of size_t is contained in the range of values of unsigned long int. But the item 3 in the list above left the door open to exceptions to this rule.

所以，我们只能说，强烈意图的实施，使得为size_t 的价值持有的 unsigned long int类型<范围/ code>。但是，我们不能完全确定。

So, we can only say that is highly intended that an implementation makes the values of size_t hold in the range of the unsigned long int. But we cannot be completely sure.

如果您要确定，你可以做以下步骤检查您的系统：

If you want to be sure, you can do the following procedure to check your system:

1. 包括文件＆LT; limits.h中＆GT; 和＆LT; stdint.h＆GT; ，以访问关于整数类型的执行情况。


2. 比较常数 ULONG_MAX （从＆LT; limits.h中＆GT; ）和 SIZE_MAX （从＆LT; stdint.h＆GT; ）。

1. Include the files <limits.h> and <stdint.h> in order to access the information about integer types for your implementation.
2. Compare the constants ULONG_MAX (from <limits.h>) and SIZE_MAX (from <stdint.h>).

一个短节目是这样的：

 #include <stdio.h>
 #include <limits.h>
 #include <stdint.h>

 int main(void) {
      printf("Is the range of size_t containd in that of unsigned long?\n\n");
      if (SIZE_MAX <= ULONG_MAX)
           printf("Yes");
      else 
           printf("No");
      return 0;
 }

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