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HappyFace
Asked: 2019-08-11 01:16:39 +0800 CST

zsh：变量无故取消设置

  • 772

v打电话后不知何故变得不安f。

$ zsh -xc 'v=1; f() { local v; v=2 true; }; f; typeset -p v'
+zsh:1> v=1
+zsh:1> f
+f:0> local v
+f:0> v=2 +f:0> true
+zsh:1> typeset -p v
zsh:typeset:1: no such variable: v

这是我的原始复制报告的要点。

我确实发送了电子邮件 [email protected]，但我还没有收到任何回复。

zsh variable

1 个回答

  1. Best Answer

    那确实是一个错误。你做了正确的事情来报告它。

    然后通过此提交修复它：https ://sourceforge.net/p/zsh/code/ci/d946f22a4cd2eed0f3a67881cfa57c805703929c/将包含在下一个版本中。

    这是维护者的解释zsh：

    On Wed, 2019-08-14 at 10:37 +0100, Stephane Chazelas wrote:
> 2019-08-08 20:38:05 +0430, Aryn Starr:
> Now, that being said, as discussed on U&L it looks like a bug
> indeed and a shorter reproducer is:
> 
> $ zsh -xc 'v=1; f() { local v; v=2 true; }; f; typeset -p v'
> +zsh:1> v=1
> +zsh:1> f
> +f:0> local v
> +f:0> v=2 +f:0> true
> +zsh:1> typeset -p v
> zsh:typeset:1: no such variable: v
> 
> Most likely, that's the "v=2 true" (where "true" is a builtin) that ends up
> unsetting the "v" from the global scope.

Yes, the saved version of "v" that we restore after the builtin is
missing the pointer back to the version of v in the enclosing scope.  So
it's not only not shown as set, it will leak memory.

This simply preserves that pointer in the copy, but this assumes we've
correctly blocked off the old parameter from being altered inside the
function scope --- if we haven't that preserved old pointer is going to
get us into trouble.  However, if we haven't that's already a bug, so
this shouldn't make things worse.

pws

    [跳过补丁]

    • 1

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