Question

There is a search and replace operation I am trying to do using backreferencing and regular expressions in vim. Interestingly, it will only recognize the pattern if I do a pure search, but if I do a search and replace it gives me an E486: pattern not found error.

I have a bunch of function calls of the form:

function( Nullable< double >(1.1), map[FOO] );

Where FOO is some different variable name on each line. I want to turn it into

function( othermap[ FOO ], map[FOO] );

If I try

:%s/Null.*\(map[\)\(.*\)\]/othermap[ \2 \], \1\2\]/g

It gives me the “Pattern not found error.” Even

:%s/Null.*\(map[\)\(.*\)\]//g 

will not work because it’s just not recognizing the pattern. But if I try the following command with the exact same search regex:

/Null.*\(map[\)\(.*\)\]

It highlights correctly. Following which, I can do %s//othermap[ \2 ], \1\2] to do my replacement. So I was able to do my replacement after all, but I can’t for the life of me understand why the pattern would be recognized in one case and not in the other.

Answer 1

I can reproduce the result using copy’n’paste from your question to my vim session. The detailed message I get, though, is:

E486: Pattern not found: Null.*\(map[\)\(.*\)\]/othermap[ \2 \], \1\2\]/g

Note that it has lost the s/ at the start.

However, looking rather carefully at this, the trouble is an unescaped [:

s/Null.*\(map[\)\(.*\)\]/othermap[ \2 \], \1\2\]/g
             ^
             |-- here; you need \[ to match the literal

I don’t use the % notation; I would automatically write:

:g/Null.*\(map\[\(.*\)\]\)/s//othermap[\2], \1/g

This has slightly different capturing. There was also no need to use the backslash in \] in the replacement string.

However, this command also works for me:

:%s/Null.*\(map\[\(.*\)\]\)/othermap[\2], \1/g