如何在VIM中将十进制数转换为十六进制？

I have this column in a file I'm editing in VIM:

我在VIM中编辑的文件中有此列:

16128
16132
16136
16140
# etc...

And I want to convert it to this column:

我想将其转换为此列:

0x3f00
0x3f04
0x3f08
0x3f0c
# etc...

How can I do this in VIM?

我怎样才能在VIM中做到这一点?

5 个解决方案

#1

Use printf (analogous to C's sprintf) with the \= command to handle the replacement:

使用printf(类似于C的sprintf)和\ =命令来处理替换:

:%s/\d\+/\=printf("0x%04x", submatch(0))

Details:

:%s/\d\+/ : Match one or more digits (\d\+) on any line (:%) and substitute (s).

:%s / \ d \ + /:匹配任何行(:%)和替换(s)上的一个或多个数字(\ d \ +)。

\= : for each match, replace with the result of the following expression:

\ =:对于每个匹配,替换为以下表达式的结果:

printf("0x%04x", : produce a string using the format "0x%04x", which corresponds to a literal 0x followed by a four digit (or more) hex number, padded with zeros.

printf(“0x%04x”,:使用格式“0x%04x”生成一个字符串,对应于文字0x后跟四位(或更多)十六进制数字,用零填充。

submatch(0) : The result of the complete match (i.e. the number).

submatch(0):完全匹配的结果(即数字)。

For more information, see:

有关更多信息,请参阅:

:help printf()
:help submatch()
:help sub-replace-special
:help :s

#2

Yet another way:

另一种方式:

:rubydo $_ = '0x%x' % $_

Or:

:perldo $_ = sprintf '0x%x', $_

This is a bit less typing and you avoid a level of quoting / shell escaping that you'd get if you did this via :!. You need Perl / Ruby support compiled into your Vim.

如果您通过以下方式执行此操作,则可以减少打字/ shell转义。您需要将Perl / Ruby支持编译到您的Vim中。

#3

another way, to pass it to awk

另一种方式,将它传递给awk

awk '{printf "0x%x\n",$1}' file

#4

Select (VISUAL) the block of lines that contains the numbers, and then:

选择(VISUAL)包含数字的行块,然后:

:!perl -ne 'printf "0x\%x\n", $_'

#5

To convert a decimal into hexadecimal, use the command

要将十进制转换为十六进制,请使用该命令

:echo printf("%x", decimal)

This results in hexadecimal equivalent of decimal being displayed.

这导致显示十六进制当量小数。

Example:

:echo printf("%x", 61444)

f004 (equivalent of 61444) is displayed.

显示f004(相当于61444)。

This will not find and replace 61444 in the file.

这将无法在文件中找到并替换61444。

#1