bash - 什么是间接扩张？ ${!var*} 是什么意思？

Question

我正在阅读“Bash 初学者指南”。它说:

If the first character of PARAMETER is an exclamation point, Bash uses the value of the variable formed from the rest of PARAMETER as the name of the variable; this variable is then expanded and that value is used in the rest of the substitution, rather than the value of PARAMETER itself. This is known as indirect expansion.

给出的例子是:

franky ~> echo ${!N*}
NNTPPORT NNTPSERVER NPX_PLUGIN_PATH

这里不太明白:

the value of the variable formed from the rest of PARAMETER

因为PARAMETER就是!N*，那么

the rest of PARAMETER

只是 N*。这怎么会形成变数呢？ Bash 是否在那里搜索了所有可能的命令？

Answer 1

如果您阅读了 bash 手册页，它基本上证实了您所说的内容:

If the first character of parameter is an exclamation point (!), a level of variable indirection is introduced. Bash uses the value of the variable formed from the rest of parameter as the name of the variable; this variable is then expanded and that value is used in the rest of the substitution, rather than the value of parameter itself. This is known as indirect expansion.

但是，从那里继续阅读:

The exceptions to this are the expansions of ${!prefix*} and ${!name[@]} described below.

${!prefix*} Names matching prefix. Expands to the names of variables whose names begin with prefix, separated by the first character of the IFS special variable.

换句话说，您的特定示例 ${!N*} 是您引用的规则的异常(exception)。但是，它确实在预期情况下如宣传的那样工作，例如:

$ export xyzzy=plugh ; export plugh=cave

$ echo ${xyzzy}  # normal, xyzzy to plugh
plugh

$ echo ${!xyzzy} # indirection, xyzzy to plugh to cave
cave