Shell Programming and Scripting

It seems that I really need some serious regexp study, in the following example,

>cat /tmp/test
abc 7878 7878 7878 7878 123 123 123 345 234 345 345 345 345

>sed 's/\([^ ]* \)*/\1/g' /tmp/test
345 345

Why sed deletes all the columns except the last two ones? From what I understood, sed should act like this

it finds "abc" match the \([^ ]* \), than there is 0 "abc" after it, so sed replaces "abc" with "abc" and continues,

why I need to add a \1 here?

sed 's/\([^ ]* \)\1*/\1/g' /tmp/test

Your first regular expression says "anything with zero or more non-spaces followed by a space" any number of times. That sounds like "a column", and that's what it means.

The second regular expression takes the "column" and says "if the same column is repeated zero or more times, replace with the first occurrence".

So "\([^ ]* *\)" can match any column anywhere; repeat it, and you match adjacent columns, with anything in them.

In contrast, \1 referst to "whatever the first group matched" -- not the regular expression, but the actual match. That's why it detects duplicates of the same column, not abstract columns.