Many quantifiers modify the character sets that precede them. Here's what you'd learn in this lesson: To have a search pattern check at the start of the beginning of a string, James shows anchoring within Regular Expressions. Using the -m option, it merges presorted input files. Regular Expressions is nothing but a pattern to match for each input line. Features of Regular Expression. I'm using BASH 3.2.48 for Mac OS X and Bash 4.1.10(4) for Cygwin (Wow, the Mac version is that old?). After all, all of the extended regular expression stuff originally came from Perl. How can I check if a directory exists in a Bash shell script? However, strings like "//////6.00007" match. Since I don't understand it, I cannot fix it; I think the basic problem I am clueless about converting valid regex from the debugger to work in bash/sed. Here's an example; look at the regex pattern carefully: Similarly, numbers in braces specify the number of times something occurs. I ran the regex in its own foo.sh as suggested by @lurker and the code ran as expected with my test cases. Finally, we look at the | (alternation) operator, which is part of the extended regex features. grep , expr , sed and awk are some of them.Bash also have =~ operator which is named as RE-match operator.In this tutorial we will look =~ operator and use cases.More information about regex command cna be found in the following tutorials. grep , expr , sed and awk are some of them.Bash also have =~ operator which is named as RE-match operator.In this tutorial we will look =~ operator and use cases.More information about regex command cna be found in the following tutorials. Actually, the best source is the Perl documentation. Easy way to test is with awk: The same C language function that awk uses to convert a string to a float is used by many other languages (Ruby, Perl, Python, C, C++, Swift, etc etc) so if you consider your format valid, you are probably going to be writing your own conversion routine as well. The following Regular Expression will match both lines because there is an empty string before word "Regular" on each line: $ grep "\