[英]OCaml: unexpected exception with Unix.getlogin when stdin redirected
I found out the next issue in this simple code: 我在这个简单的代码中找到了下一个问题:
let () =
print_endline "Hello";
print_endline (Unix.getlogin ())
Running in the normal case, with ./a.out
gives: 在正常情况下运行,使用
./a.out
给出:
Hello
ricardo
But running like ./a.out </dev/null
makes Unix.getlogin fail: 但是像
./a.out </dev/null
一样运行会使Unix.getlogin失败:
Hello
Fatal error: exception Unix.Unix_error(20, "getlogin", "")
Any idea why this happens? 知道为什么会这样吗?
Redirecting the input of a program overrides its controlling terminal. 重定向程序的输入会覆盖其控制终端。 Without a controlling terminal, there is no login to be found:
没有控制终端,就找不到登录信息:
$ tty
/dev/pts/2
$ tty < /dev/null
not a tty
You can, however, still find a user's name (perhaps) by getting the user's id ( getuid
) and looking up his passwd entry (related docs) ( getpwuid
), then finding his username in it. 但是,您仍然可以通过获取用户的id(
getuid
)并查找其passwd条目(相关文档) ( getpwuid
),然后在其中查找用户名来查找用户名(可能)。
Depending on your application: 取决于您的应用:
if you don't really care about the value returned by "getlogin", you can do something like: 如果你真的不关心“getlogin”返回的值,你可以这样做:
try Unix.getlogin () with _ -> Sys.getenv "USER"
you will probably get something better than getuid
, since it will also work for programs with Set-User-ID flags (sudo/su). 你可能会得到比
getuid
更好的东西,因为它也适用于具有Set-User-ID标志(sudo / su)的程序。
if you really care about the value returned by "getlogin", ie you really want to know who is logged in, you should just fail when getlogin fails. 如果你真的关心“getlogin”返回的值,即你真的想知道谁登录了,你应该在getlogin失败时失败。 Any other solution will give you only an approximation of the correct result.
任何其他解决方案只会给出正确结果的近似值。
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