表达式中涉及数学除法时，如何使用vim regex替换文本 [英] How can I use vim regex to replace text when math divide is involved in the expression

问题描述

我正在使用vim处理如下文本

I am using vim to process text like the following

0x8000   INDEX1 ....
0x8080   INDEX2 ....
....
0x8800   INDEXn ....

我想使用正则表达式获取每行的索引号.那是

I want to use regular expression to get the index number of each line. that is

0x8000 ~ 0
0x8080 ~ 1
....
0x8800 ~ n

数学评估应为(hex-0x8000)/0x80.我正在尝试使用vim正则表达式替换来获取结果

The math evaluation should be (hex - 0x8000) / 0x80. I am trying to using vim regular expression substitution to get the result in line

%s/^\(\x\+\)/\=printf("%d", submatch(1) - 0x8000)

这将产生

0     INDEX0
128   INDEX1
....
2048  INDEXn

我想要做的是将其进一步更改为

What I want to do is to further change it to

0     INDEX0
1     INDEX1
...
20    INDEXn

也就是说，我想用0x80进一步划分第一列.这是我遇到问题的时候.

That is, I want to further divide the first column with an 0x80. Here is when I get the problem.

原始参数为"submatch(1)-0x8000".我现在在其中添加一个"/0x80"，它形成

The original argument is "submatch(1) - 0x8000". I now add an "/ 0x80" to it, which forms

%s/^\(\x\+\)/\=printf("%d", (submatch(1) - 0x8000)\/0x80)

现在Vim报告错误

Invalid expression: printf("%d", (submatch(1) - 0x8000)\/0x80))

处理"/"时似乎vim遇到问题.我也尝试使用单个"/"(不进行转义)，但仍然失败.

It looks like vim meet problem when processing "/". I also tried with a single "/" (without escape), but still fails.

有人可以帮我吗?

推荐答案

您不能在sub-replace-expression中使用分隔字符.
从 :h sub-replace-expression:

You can't use the separation character in a sub-replace-expression.
From :h sub-replace-expression :

Be careful: The separation character must not appear in the expression!
Consider using a character like "@" or ":".  There is no problem if the result
of the expression contains the separation character.

---

相反，将分隔符更改为不再与除法运算符匹配.例如，使用#.

:%s#^\(0x\x\+\)#\=printf("%d", (submatch(1) - 0x8000)/0x80)

请注意，我必须更改您的正则表达式(特别是将^\(\x\+\)更改为^\(0x\x\+\)).我不知道您为什么要为您工作，但是从:h character-classes开始，\x不应包含尾随的0x:

Note that I had to change your regex (specifically ^\(\x\+\) to ^\(0x\x\+\)). I don't know why yours works for you, but from :h character-classes, \x shouldn't include the trailing 0x :

/\x     \x      \x      hex digit:                      [0-9A-Fa-f] 

---

此外，使用非常魔术的模式(请参见:h magic)，您的正则表达式(至少对我来说)更容易阅读:

---

Also, your regex is a bit easier to read (to me at least) using very-magic mode (see :h magic):

:%s#\v^(0x\x+)#\=printf("%d", (submatch(1) - 0x8000)/0x80)

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