c - fgetc，检查 EOF

Question

在 Linux System Programming 这本书中，我读过一些这样的书:

fgetc returns the character read as an unsigned char cast to an int or EOF on end of file or error. A common error using fgetc is:

char c;
if ((c = fgetc()) != EOF) {...}

The right version of this code is:

int c;
if ((c = fgetc()) != EOF) { printf("%c", (char)c); ... }

那么，为什么我不能在与 EOF 比较之前将返回值转换为 char？为什么我必须将 EOF 与 int 进行精确比较？由于 EOF 定义为 -1，它不是通常被转换为 char 吗？
有平台/编译器吗？真的吗？

Answer 1

不能将返回值转换为 char，因为返回值可能是 EOF，而 EOF 值是系统相关的，不等于任何有效的字符代码. link

通常它是 -1 但您不应该这样假设。

查看来自 c-faq-site 的出色答案:

Two failure modes are possible if, as in the fragment above, getchar's return value is assigned to a char.

1. If type char is signed, and if EOF is defined (as is usual) as -1, the character with the decimal value 255 ('\377' or '\xff' in C) will be sign-extended and will compare equal to EOF, prematurely terminating the input. (assuming 8 bits char).

2. If type char is unsigned, an actual EOF value will be truncated (by having its higher-order bits discarded, probably resulting in 255 or 0xff) and will not be recognized as EOF, resulting in effectively infinite input.

希望对您有所帮助!

已编辑:(在这个答案上添加了@FatalError 评论，这在 c-faq 网站上有解释，但这对我来说看起来更清楚)

“如果将其转换为 char，则 EOF 将采用与某个有效字符相同的值，因此与该字符无法区分。仅此一项就足以证明不会将结果设为 char”@FatalError 评论。