潜在的问题与C标准malloc'ing字符 [英] Potential problem with C standard malloc'ing chars

问题描述

在回答了评论我的另一个答案<一个href=\"http://stackoverflow.com/questions/1247964/how-do-you-know-how-much-space-to-allocate-with-malloc/1247977#1247977\">here,我发现我的想法的可能的是C标准的孔（C1X，我没有检查与先前的，是的，我知道这是令人难以置信的可能性不大，我独自各个星球的居民中已经发现了在标准的bug）。信息如下：

When answering a comment to another answer of mine here, I found what I think may be a hole in the C standard (c1x, I haven't checked the earlier ones and yes, I know it's incredibly unlikely that I alone among all the planet's inhabitants have found a bug in the standard). Information follows:

1. 6.5.3.4节（sizeof操作符）第2条规定sizeof操作符得到其操作数的大小（以字节为单位）。


2. 该节第3段规定：当应用于具有char类型，unsigned char型，或符号的字符，（或其合格的版本）的结果为1的操作数。


3. 第7.20.3.3描述的void * malloc的（为size_t SZ），但所有这说的是malloc函数为对象，其大小分配空间由size指定，其值是不确定的。它并没有提及在使用所有什么单位的说法。


4. 附件E startes 8是的最小的为 CHAR_BIT 值，使字符的长度可以是多个字节。

1. Section 6.5.3.4 ("The sizeof operator") para 2 states "The sizeof operator yields the size (in bytes) of its operand".
2. Para 3 of that section states: "When applied to an operand that has type char, unsigned char, or signed char, (or a qualified version thereof) the result is 1".
3. Section 7.20.3.3 describes void *malloc(size_t sz) but all it says is "The malloc function allocates space for an object whose size is specified by size and whose value is indeterminate". It makes no mention at all what units are used for the argument.
4. Annex E startes the 8 is the minimum value for CHAR_BIT so chars can be more than one byte in length.

我的问题很简单：

在其中char是16位宽，环境将的malloc（10 * sizeof的（字符））拨款1000个字符（20字节）或10个字节？上述第1点似乎表明前者，点2表示后者。

In an environment where a char is 16 bits wide, will malloc(10 * sizeof(char)) allocate 10 chars (20 bytes) or 10 bytes? Point 1 above seems to indicate the former, point 2 indicates the latter.

任何人有更多的C-标准福比我有这样一个答案？

Anyone with more C-standard-fu than me have an answer for this?

推荐答案

在16位字符环境的malloc（10 * sizeof的（CHAR ））将拨出10 字符 S（10字节），因为如果字符 16位，则该架构/实现定义一个字节为16位。 A 字符不是一个八位位组，这是一个字节。在旧的计算机上这可能会比8位的事实上的的标准，我们有今天。

In a 16-bit char environment malloc(10 * sizeof(char)) will allocate 10 chars (10 bytes), because if char is 16 bits, then that architecture/implementation defines a byte as 16 bits. A char isn't an octet, it's a byte. On older computers this can be larger than the 8 bit de-facto standard we have today.

从C标准的有关部分如下：

The relevant section from the C standard follows:

3.6术语，定义和符号

3.6 Terms, definitions and symbols

字节 - 数据存储足以容纳基本字符集执行环境的...

byte - addressable unit of data storage large enough to hold any member of the basic character set of the execution environment...

注2 - 一个字节由比特的连续序列的，其数量是实现定义

NOTE 2 - A byte is composed of a contiguous sequence of bits, the number of which is implementation-defined.

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