lappend varName ?value value ...?
http://www.purl.org/tcl/home/man/tcl8.4/TclCmd/lappend.htm
Appends each value as a list element onto the end of the list stored in varName.
lappend used to be the only list-command which requires the name of a variable containing a list rather than the list itself, but in Tcl 8.4 the lset command was introduced.
lappend listName
rather than
lappend $listName
This command differs from append in that values are appended as list elements rather than raw text.
lappend will not make varName a list ; if it is one before the command, it will continue to be one.
AJD Uh? lappend *will* attempt to make varName's value into a list. The docs say "This command treats the variable given by varName as a list." ie. if it isn't a list to begin with, tcl will attempt to convert it to one, giving an error this is not possible:
% set x "a \{" ;# string
a {
% lappend x b ;# treat it as list
unmatched open brace in list ;# it wasn't a proper list!LV Actually, the example here of x and b demonstrates that Tcl does not try to make varName into a list, but instead presumes that it is a valid list. If it attempted to turn the variable into a list, it would surely succeed; in this case, however the basic assumption of (nearly?) all the l* commands is that list arguments are proper lists and that no attempt will be made to second guess the caller.
Lars H: It seems you disagree about what it means to "attempt" something. What lappend does is that it attempts to generate a list internal representation for the current value of the variable (i.e., it calls tclListType.setFromAnyProc [L1 ]), and then appends the element(s) to that. This attempt will fail when the string representation cannot be interpreted as a list.
LV Actually, what I was disagreeing with was the idea the phrase "make varName's value into a list". This was based on my interpretation of the docs - which was a mistake on my part. However, user of lappend needs to understand that lappend will not turn a string value into a list; if the string is not a list, then a call to lappend fails. And that's really the bottom line of the discussion. Sorry for the digression.
You do not need to assign the results of lappend to varName; lappend modifies varName directly.
How would one append an item to the beginning of a list? Since append means to the end of, one wouldn't append to the beginning of a list. Instead, for a way to perform the equivalent of prepend, see linsert -- jfr
LV [tried proposing a couple ways to prepend, all of which failed... RS [provided the comment that...]]]
set a [concat $toPrepend $a]
results in the 4-element list (maybe via shimmering over string reps)
xyz 1 2 3
linsert would keep an existing list representation, so I presume its faster to do
set a [linsert $a 0 $toPrepend]
RS 2006-10-01: jcw contributed this (for me) surprising idiom:
lappend var
does nothing, but creates var if it doesn't exist. Similar to, but simpler than:
if {![info exists var]} {set var ""}This may however fail if $var holds a string that cannot be parsed as a list.
See also list, lindex, linsert, llength, lrange, lreplace, lsearch, lsort .
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