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emacs regex help

开发者 https://www.devze.com 2023-02-01 22:10 出处:网络
Okay, this should be s开发者_StackOverflow中文版uper easy, I must be doing something dumb. I\'ve got this:

Okay, this should be s开发者_StackOverflow中文版uper easy, I must be doing something dumb.

I've got this:

double foo3;

I want to replace it with:

archiver & BOOST_SERIALIZATION(foo3);

I do this:

M-x query-replace-regxp RET double \(a-z0-9+\) RET archiver & BOOST_SERIALIZATION(\1)

Emacs says 0 matches.


You mist the List Operator:

double \([a-z0-9]+\)
0

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