实在忍不住了,一直想不通的问题, backslash在$() 和 ``中的不同

yongjian

Post by johnny_jiang
But why $x is substituted for its value 1, I think '$' should be escaped by '\' in ``.

no, a single "\" in `` means nothing. it will get rid of it first and then bring the rest to subshell. if you want to get $x, you need to do

1. echo `echo \\$x`

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, so that `` will treat it as its literal value "\" and bring it to the subshell, in that case, bash will see "echo \$x", which is the same as $(echo \$x). In bash echo, a single "\" will be treated literally, so it can esc "$".

johnny_jiang

Post by yongjian
no, a single "\" in `` means nothing. it will get rid of it first and then bring the rest to subshell. if you want to get $x, you need to do

1. echo `echo \\$x`

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, so that `` will treat it as its literal value "\" and bring it to the subshell, in that case, bash will see "echo \$x", which is the same as $(echo \$x). In bash echo, a single "\" will be treated literally, so it can esc "$".

yongjian兄：

鉴于怕产生理解上的错误，我还是用中文表达吧。

no, a single "\" in `` means nothing.

以上这句是指\将不被当前shell作为metacharacter处理对吗，所以不逃逸后面的$?

In bash echo, a single "\" will be treated literally, so it can esc "$".

对上面这句不理解，单个\被最为普通字符处理，而又可以逃逸$？

这样的话，问题就有两个：
1。如果命令替换中存在backslash，它是在子shell中被处理的吗，会对其后紧跟的特殊字符（例如$） 产生逃逸作用呢？
2。根据大家的总结，在``中\被特殊处理，而在$()中\被认为普通字符

忘yongjian兄不惜赐教。小弟是个刨根问底的人，麻烦了！！！:thank

yongjian

simply, in ``, "\" is a metadata, so `` will process it first before the cmd gets sent to subshell. In order to let subshell see one "\", you will have to have "\\" in ``. A single "\" in `` will simply be ignored. Where in $(), the situation is different, $() treats "\" literally, so it sends whatever in $() to the subshell, so only one "\" is enough.

johnny_jiang

To yongjian:

From your statement, does it mean `` will treat \ in current shell before it gets sent to the subshell?
If it is, when the parameter substitution happens?
I mean

1. 
   
2. bash# x=1
   
3. bash# echo `echo \$x`
   
4. bash# 1
   

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When is $x replaced, or rather, where is $x replaced? In current shell or subshell?

I always take it that $x is replaced in current shell and before command substitution because parameter expansion is ahead of command substitution.

So I'm confued why $x is not influenced by the preceding \.

yongjian

I don't think bash or any other shells will take apart the `` or $(), instead, it will do the cmd sub first and take whatever inside to subshell. Since $x is inside ``, so it shouldn't be processed before the cmd sub but after, which is actually in subshell.

johnny_jiang

Post by yongjian
I don't think bash or any other shells will take apart the `` or $(), instead, it will do the cmd sub first and take whatever inside to subshell. Since $x is inside ``, so it shouldn't be processed before the cmd sub but after, which is actually in subshell.

But $x is not an exported variable, that is, it is not an environmental variable.

Subshell will inherit it?

wplxb

Post by johnny_jiang
But $x is not an exported variable, that is, it is not an environmental variable.

Subshell will inherit it?

建议参考一下36贴

另：能输入汉字的话就用中文回贴吧，为了让大家更容易看懂，而且毕竟这里是中文论坛；yongjian可能只是无法输入汉字。
还有，escape应该翻译成“转义”而不是“逃逸”吧

苦涩之恋

好帖，我打算在找到工作后学习SHELL

johnny_jiang

Post by wplxb
建议参考一下36贴

另：能输入汉字的话就用中文回贴吧，为了让大家更容易看懂，而且毕竟这里是中文论坛；yongjian可能只是无法输入汉字。
还有，escape应该翻译成“转义”而不是“逃逸”吧

呵呵，多谢wplxb的建议

有些地方的翻译为逃逸，其实意思是一样的，也许我引用的不正确吧
36贴我看过了，难道不是环境变量的变量shell也能继承吗？我应该不行吧。

yongjian

Post by johnny_jiang
呵呵，多谢wplxb的建议

有些地方的翻译为逃逸，其实意思是一样的，也许我引用的不正确吧
36贴我看过了，难道不是环境变量的变量shell也能继承吗？我应该不行吧。

subshell inherits everything from parent shell but not the other way, unless you use export..

1. a=1;echo `echo $a`

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an easy sample. For using en... I was able to type ch but several weeks ago, I upgraded to dapper, and firefox just crashes every time when I use it. I switched to opera but it doesn't support ch input... have no choice right now but waiting for a stable firefox update.