Regex for searching in source code( while loop)

Question

I want to create regular expression which matches while(true) loop in which body is not used word 'argv'

Example 1:

    while(true){ 
        for(int i = 0; i < argc; i++) printf("%s",argv[i]);
    }

argv is used in the body of while loop - reg ex should not match this

Example 2:

 while(true){ 
     for(int i = 0; i < 5; i++) printf("%d",i);
 }

argv is not used in the body of while loop - reg ex should match this.

Till now i have just this regular expression:

while\s*\(\s*true\s*\)\s*\{ there_should_be_something \}

I dont know what to replace with there_should_be_something

Thanks a lot

Using regular expressions to parse non-regular languages like C can be rather difficult and non-rewarding. Consider using an actual parser. — unwind
– unwind, Commented Apr 13, 2017 at 13:23

Tamas Rev · Accepted Answer · 2017-04-13 14:03:26Z

1

Here is a regex that works on the examples above: while\s*\(\s*true\s*\)\s*\{(([^}](?!argv))*)\}

Explanation:

while\s* matches while, then zero or more spaces
\(\s*true\s*\) matches (true) and tolerates spaces
\{...\} tolerates stuff between the braces
The stuff within the braces is based on this: [^}]*, i.e. no braces are allowed within the braces. This is a limitation for the sake simple regex. If you want to deal with all the braces, I suggest to sue a real parser.
This stuff within the braces is combined with negative lookahead: (?!argv) which means anything that's not followed by argv.
I.e. [^}](?!argv) means any non-brace character that's not followed by argv.
We just group it and require this group zero or more times: ([^}](?!argv))*

answered Apr 13, 2017 at 13:44

Tamas Rev

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