How to do the following replace in vim?

Question

I want to replace all of the occurrences of

armv7_format_strings<any_string> = {<str1>, <str2>};

by

armv7_format_strings<any_string>[0] = <str1> ;
armv7_format_strings<any_string>[1] = <str2> ;

Answer 1

This one worked:

%s/\(armv7_format_strings.*\) = {\(.*\), \(.*\)};/\1[0] = \2;\r\1[1] = \3;\r/gc

Answer 2

You can use a substation with a sub-replace-expression.

:%s/^\(\s*armv7_format_strings.*\) = {\([^}]*\)};/\=join(map(split(submatch(2), '\s*,\s*'), 'submatch(1)."[".v:key."] = ".v:val.";"'),"\r")/

Overview

Basically this captures the left hand side ({lhs}) and the right hand side ({rhs}) separately. Then splits the {rhs} by , (aka parameter) and concatenates the {lhs} to each new parameter. Finish the substitution by joining each new line together with a new line, "\r".

Note: this will fail for tricky strings, e.g. "foo,bar" as this method just assumes no , or } inside of strings.

Glory of Details

• %s:^\(\s*armv7_format_strings.*\) = {\([^}]*\)};/\={expr}/ - Capture the {lhs} and {rhs}
• use an sub-replace-expression, \={expr}, to use {expr} as the replacement.
• submatch(1) is the {lhs} aka \1 in normal regex speak
• submatch(2) is the {rhs} aka \2 in normal regex speak
• split(submatch(2), '\s*,\s*') - split the {rhs} by ,. Yields are parameters as a List
• map({lst}, {expr}) - map() will update the list by evaluating {expr}
• use submatch(1)."[".v:key."] = ".v:val.";" as our expression, {expr}
• v:val will the each single item in the List
• v:key will the the index of the List
• join({lst}, "\r") will join the list together as new lines

For more help

:h sub-replace-expression
:h join(
:h map(
:h split(
:h submatch(

Answer 3

While I'd use the regex swap you could also do it with a macro.

qq^vt=h"ayf{lvt,"byf,lvt}"cydd<CTRL_r>a[0] = <CTRL_r>b;<Enter><CTRL_r>a[1] = <CTRL_r>c;q

Its basically the same as the regex. Yank the left side of the equals into register a, and the two strings in the array into registers b and c. Delete the line, and paste RegA[0] = RegB;\nRegA[1] = RegC;