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regexlinuxcommand-linegrep
Lee Torres
Lee Torres

Reputation: 561

Grep: Finding in which functions a function is called

Let's say I have a folder with numerous modules, and I'm doing a grep for a function. Let's say the function I'm searching for is called test(). Is it possible to find out which functions test() is called in? For example, let's say below is ModuleA.erl, and we do a grep for test().

FunctionA()
   A = 5, 
   B = 3,
   test()

It should return:

ModuleA.erl: FunctionA()

It is safe to assume that the function definition will always be non-indented

Upvotes: 1

Views: 584

Answers (2)

Avinash Raj
Avinash Raj

Reputation: 174706

Just run the below grep command on the directory where the .erl files are stored.

grep -HPoz '^\S.*(?=(\n .*)+test\(\))' *.erl

From man grep

-H, --with-filename       print the file name for each match
-P, --perl-regexp         PATTERN is a Perl regular expression
-z, --null-data           a data line ends in 0 byte, not newline

^ asserts that we are at the start. \S matches any non-space character..* matches any character but not of line breaks, zero or more times. (?=(\n .*)+test\(\)) Positive lookahead assertion which asserts that the match must be followed by

  • \n .* A new line character and a space followed by any number of characters,

  • (\n .*)+ one or more times.

  • test\(\) and must have a test() string. Match will occur only if this condition was satisfied.

Test:

Folder$ cat gi.erl
FunctionA()
       A = 5, 
       B = 3,

FunctionB()
       A = 5, 
       B = 3,
       test()
Folder$ cat ri.erl
FunctionA()
   A = 5, 
   B = 3,
   test()
FunctionB()
   A = 5, 
   B = 3,
Folder$ grep -HPoz '^\S.*(?=(\n .*)+test\(\))' *.erl
gi.erl:FunctionB()
ri.erl:FunctionA()

Here Folder is the folder name where all the erl files are stored.

Upvotes: 1

fedorqui
fedorqui

Reputation: 289585

You can use awk for this:

$ awk -v patt="test" '$0 ~ /^\w/ {f=$0} $0 ~ patt {print f}' file
FunctionA()

Explanation

This stores the lines not starting with a space. If a line matches, then it prints the last stored line.

  • -v patt="test" provide the pattern
  • $0 ~ /^\w/ {f=$0} if the current line starts with some character (not space), store its value.
  • $0 ~ patt {print f} if a line contains the given pattern, print the stored header.

Test

$ cat a
FunctionA()
   A = 5, 
   B = 3,
   test()
FunctionB()
   A = 5, 
   B = 3,
   another()
$ awk -v patt="test" '$0 ~ /^\w/ {f=$0} $0 ~ patt {print f}' a
FunctionA()

Upvotes: 3

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