tcl如何在正则表达式输出中处理方括号？

Question

I try to use regular expression command like:

IN: regexp -all -inline {\S+} "RR\[0\] in"
OUT:{RR[0]} in

But,when there is no square bracket in the string,the output format is different.

IN: regexp -all -inline {\S+} "RR in"
OUT:RR in

Why does the first element is between curly braces in the first case?

Answer 1

The issue is the square brackets are reserved in tcl for executing commands.

So it evaluate every square brackets found, included those inside strings delimited by " .

So when it returns the regexp matching it put the curling brackets to protect the string to be interpreted as containing a command.

Even that, the results continue to be a string.

For example if you do that:

% set r [regexp -all -inline {\S+} "RR\[0\] in"]
{RR[0]} in

And then you print from a loop:

% foreach x $r { puts $x }
RR[0]
in

By the join command you could easily convert any kind of list into a string when you need, avoiding to remove the brackets with trim functions.

% puts [join $r ]
RR[0] in

or direct to the element:

% puts [join [lindex $r 0] ]
RR[0]

Answer 2

First, there is an error when you run the regexp the way it is. It should be:

regexp -all -inline {\S+} {RR[0] in}

With braces around the string you search in. Because the square barckets are interpreted as a command, if you must tell tcl not to make a substitution, and one of the ways to tell it is by using braces.

Saying that, when the brackets do not exist, the regexp won't find them, of course.