什么时候在C中无符号和有符号字符指针之间的转换变得不安全？ [英] When does conversion between unsigned and signed character pointer becomes unsafe in C?

问题描述

如果我在 clang 和 Visual Studio 中都这样做：

If I do this in both clang and Visual Studio:

  unsigned char *a = 0;
  char * b = 0;

  char x = '3';

  a = & x;

  b = (unsigned char*) a;

我得到警告，我正在尝试在有符号和无符号字符指针之间进行转换，但是代码确实有效。尽管编译器之所以这样说是有原因的。您能指出可能会变成问题的情况吗？

I get the warning that I am trying to convert between signed and unsigned character pointer but the code sure works. Though compiler is saying it for a reason. Can you point out a situation where this can turn into a problem?

推荐答案

为了简单起见，因为 char 表示：

To make it very simple because char represents:

• 单个字符（ char ，无论是否签名都没有关系。当您分配诸如'A'之类的字符时，您要做的是在该内存位置中写入 A ASCII码（65）。


• 一个字符串（当用作 char 缓冲区的数组或指针时）。


• 八个数字（带或不带符号）。

• A single character (char, it doesn't matter if signed or not). When you assign a character like 'A' what you're doing is to write A ASCII code (65) in that memory location.
• A string (when used as array or pointer to a char buffer).
• An eight bit number (with or without sign).

然后，当您将诸如-1的带符号字节转换为无符号字节会丢失信息（至少有符号但也可能有数字），这就是为什么您会收到警告：

Then when you convert a signed byte like -1 to unsigned byte you'll loose information (at least sign but probably number too), that's why you get a warning:

signed char a = -1;
unsigned char b = (unsigned char)a;
if ((int)b == -1)
    ; // No! Now b is 255!

值可能不是，但如果系统不代表则为1具有2的补数的负数，在该示例中，它并不重要（我从未使用过类似的系统，但它们确实存在），因为概念是有符号/无符号转换可能会丢弃信息。不管是由于显式强制转换还是通过指针强制转换而发生，这都没有关系：位将表示其他内容（结果将根据实现，环境和实际值而改变）。

Value may not be 255 but 1 if your system doesn't represent negative numbers with 2's complement, in that example it doesn't really matter (and I never worked with any system like that but they exist) because the concept is a signed/unsigned conversion may discard information. It doesn't matter if this happens because of an explicit cast or a cast through pointers: bits will represent something else (and result will change according to implementation, environment and actual value).

请注意，对于C标准 char ，带符号的字符和无符号的字符是形式上不同的类型。您将不会在乎（VS会将 char 默认为 signed 或 unsigned ，但这不是可移植的），您可能需要强制转换。

Note that for C standard char, signed char and unsigned char are formally distinct types. You won't care (and VS will default char to signed or unsigned according to a compiler option but this isn't portable) and you may need casting.

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