Different version of tcl giving different answers

Question

I am running this function in tcl 8.0 verison.

proc test {} {
    set owner 212549316
    set val [regexp {[0-9]{9}} $owner]
    puts $val
}

The same code in tcl 8.6, the output is 1 but in tcl 8.0 it is 0. I am checking whether the string contains only 9 digits in tcl 8.0.

Any help how to make it work in tcl 8.0 verison.

Interesting. Does `[0-9]\{9\}` produce expected result in Tcl 8.0? From what [I see](http://www.wellho.net/regex/tcl.html), limiting quantifiers have been introduced in Tcl with version 8.1. — Wiktor Stribiżew, May 30 '16 at 09:56
@WiktorStribiżew this is giving 0 in both versions of tcl i.e 8.0 & 8.6. — Ashish, May 30 '16 at 09:59
So, the limiting (or *bound*) quantifier just is not supported in Tcl 8.0 — Wiktor Stribiżew, May 30 '16 at 10:01

score 4 · Accepted Answer · answered May 30 '16 at 10:05

In Tcl 8.0, bound (or limiting) quantifiers are not supported.

To match 9 digits in Tcl 8.0, you'd have to repeat the [0-9] 9 times:

set val [regexp {[0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9]} $owner]

Bound quantifiers are supported starting with Tcl 8.1 with the introduction of advanced regular expression syntax.

Basic regex syntax available in Tcl 8.0 only includes:

.    Matches any character.
*    Matches zero or more instances of the previous pattern item.
+    Matches one or more instances of the previous pattern item.
?    Matches zero or one instances of the previous pattern item.
( )  Groups a subpattern. The repetition and alternation operators apply to the preceding subpattern.
|    Alternation.
[ ]  Delimit a set of characters. Ranges are specified as [x-y]. If the first character in the set is ^, then there is a match if the remaining characters in the set are not present.
^    Anchor the pattern to the beginning of the string. Only when first.
$    Anchor the pattern to the end of the string. Only when last.

Different version of tcl giving different answers

1 Answers1