“expr match”正则表达式中的“空格”字符有什么特别之处？

Question

In a bash shell, I set line like so:

line="total active bytes:         256"

Now, I just want to get the digits from that line so I do:

echo $(expr match "$line" '.*\([[:digit:]]*\)' )

and I don't get anything. But, if I add a space character before the first backslash in the regexp, then it works:

echo $(expr match "$line" '.* \([[:digit:]]*\)' )

Why?

Answer 1

The space isn't special at all. What's happening is that in the first case, the .* matches the entire string (ie, it matches "greedily"), including the numbers, and since you've quantified the digits with * (as opposed to \\+ ), that part of the regex is allowed to match 0 characters.

By putting a space before the digit match, the first part can only match up to but not including the last space in the string, leaving the digits to be matched by \\([[:digit:]]*\\) .