是将uid_t类型符号或无符号？ [英] Is the uid_t type signed or unsigned?

问题描述

我知道，标准不说关于的符号性东西将uid_t 或 gid_t 。

不一致：

网页。 http://www.gnu.org/software /libc/manual/html_node/Reading-Persona.html 说：

  

在GNU C库，这是对的无符号整型的别名。

但男人setreuid 说：

  

直供值 1 作为无论是真实或有效用户ID强制系统离开该ID不变。

问题：

1. 那么，是将uid_t GNU库的符号或无符号？




2. 我如何能提供 1 如果将uid_t 和 gid_t 是无符号的（ 1 将被转换为 0xFFFFFFFF的）？

解决方案

将uid_t 是（经过一些类型定义/定义）定义为 __ U32_TYPE 被定义为 unsigned int类型（这是我的Gentoo Linux系统）。

但是，仅仅因为 1 有特殊的意义并不意味着UID将被限制为适合在签署 INT <号码/ code>。它只是意味着最高值（即（无符号整数）-1 ）是不是一个有效的UID。在 setreuid 的code可能使用该投的反向形式（（符号的int）RUID ）比较虽然它接受对 1 干净的将uid_t 。

I know that the standard doesn't say anything about the signedness of uid_t or gid_t.

Inconsistency:

Page http://www.gnu.org/software/libc/manual/html_node/Reading-Persona.html says:

In the GNU C Library, this is an alias for unsigned int.

But man setreuid says:

Supplying a value of -1 for either the real or effective user ID forces the system to leave that ID unchanged.

Questions:

1. So, is uid_t signed or unsigned in the GNU Library?

2. How can I supply -1 if uid_t and gid_t are unsigned (-1 will be converted to 0xFFFFFFFF)?

解决方案

uid_t is (after some typedefs/defines) defined as __U32_TYPE which is defined as unsigned int (that is on my Gentoo Linux system).

However, just because -1 has a special meaning it does not mean that UIDs are restricted to the numbers that fit in a signed int. It just means that the highest value (i.e. (unsigned int)-1) is not a valid UID. The code in setreuid probably uses the reverse form of that cast ((signed int)ruid) to compare against -1 cleanly although it accepts an uid_t.

这篇关于是将uid_t类型符号或无符号？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！