Tips for VIM replacement

1. Convert Enum in C language to Java constant:

G/^ \ s * $/d
G/^/S // public static final int/

Leti = 1000
G/,/S // \ = '. I.'; '/| let I = I + 1

2. # define macro in C to convert Java constants:

% S/^ # define [A-Z, 0-9, _] * \ Zs \ s \ ze/=/
G/# define/S // public static final int/
G/$/S //;/

 

 

Some advanced replacement techniques in VIM

1. Replace Variables
In a regular expression, use the \ (and \) symbol to enclose a regular expression. Then, use variables such as \ 1 and \ 2 to access the content in \ (and.
Example:
· Change data1 data2 to data2 data1
-----------------------------------
: S/\ (\ W \ + \) \ s \ + \ (\ W \ + \)/\ 2 \ t \ 1
-----------------------------------
· Change Doe and John to john doe
-----------------------------------
: % S/\ (\ W \ + \), \ (\ W \ + \)/\ 2 \ 1/
-----------------------------------
2. Use Function replacement
Usage:
---------------------------
: S/replacement string/\ = Function Type
---------------------------
Note:
· There can be multiple function types, and the return value can be a string connector. connect, such as line (". ") returns the matched row number (: Help Line (). submatch (n) can reference the content of \ 1 and \ 2. submatch (0) references the entire content of the matching;
· The function can also be a String constant, which is caused by double quotation marks. Function expressions can also be arbitrary expressions that need to be caused by parentheses, such as (3 + 2*6 );
· The function type can also be the content in the register, accessed through "@ register name", such as @ A (no quotation marks are required, but still need to be connected using );
Example:
· Convert the following eight rows of data0 to data0 in sequence ~ 7 (the preceding number is the row number)
Double data0;
Double data0;
Double data0;
Double data0;
Double data0;
Double data0;
Double data0;
Double data0;
If Perl is installed, I used Perl to do this:
---------------------------------------------------------------------------
: R! Perl-n-Le "If (S/^ (double \ s + Data) \ D;/$1 $ K;/) {$ K ++; print $ _;} "%: P ---------------------------------------------------------------------------
%: P indicates the file name that contains the complete path.
Now we can do this with vim:
First, move the cursor to line 125 (line (".") to return the row number of the current matched row, line ("'a") to return the row number of Mark ):
---------------------------------------------------------------------------
Ma
: % S/^ \ (double \ s \ + data \) \ D \ (; \)/\ = submatch (1 ). (line (". ")-line (" 'a ") + 1 ). submatch (2)/g
---------------------------------------------------------------------------
Or
First, move the cursor to line 125 ("'v7j
: % S/^ \ (double \ s \ + data \) \ D \ (; \)/\ = submatch (1 ). (line (". ")-line ("'
---------------------------------------------------------------------------
3. start and end of matching
\ Zs and \ ze can be used to specify the start and end of replacement in the replacement operation. In the above example, they can be implemented with shorter operations:
---------------------------------------------------------------------------
Ma
: % S # ^ double \ s \ + data \ Zs \ D \ ze # \ = (line (".")-line ("'a") + 1) # G
---------------------------------------------------------------------------
In addition, you can use the VIM expression to implement the above functions:
---------------------------------------------------------------------------
: Let n = 0 | G/^ double \ s \ + data \ Zs \ D/S // \ = N/| Let N + = 1
---------------------------------------------------------------------------
Where | is used to separate unused commands;
· G executes the specified ex command in the line matching the following pattern
· \ Zs indicates that the matching starts from this
· S // \ = N: Replace the matching mode. If the matching mode is null, the matching mode is successful,
It also indicates that the replacement is an expression;
4. Advanced incremental replacement
Put the following sentence in _ vimrc
-------------------------------
Let G: I = 0
Function! INC (increment)
Let G: I = G: I + A: Increment
Return G: I
Endfunction
------------------------------
· The preceding problems can be solved as follows:
-----------------------------------------------------
: Let I =-1 | % S/^ double \ s \ + data \ Zs \ D \ ze/\ = Inc (1 )/
-----------------------------------------------------
· You can also generate digital sequences, such as generating 1 ~ A numerical sequence separated by 5 at 100:
--------------------------------------
: Let I = 0 | 'a, 'B S/^/\ = Inc (5 )/
--------------------------------------
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