Version 13 of lreplace

Replace elements in a list with new elements

Eventually, the documentation for this should appear at http://www.purl.org/tcl/home/man/tcl8.5/TclCmd/lreplace.htm

 lreplace list first last ?element element ...?

Lreplace returns a new list formed by replacing one or more elements of list with the element arguments. First gives the index in list of the first element to be replaced (0 refers to the first element). If first is less than zero then it refers to the first element of list; the element indicated by first must exist in the list. Last gives the index in list of the last element to be replaced. If last is less than first then no elements are deleted; the new elements are simply inserted before first. First or last may be end (or any abbreviation of it) to refer to the last element of the list.

The element arguments specify zero or more new arguments to be added to the list in place of those that were deleted. Each element argument will become a separate element of the list. If no element arguments are specified, then the elements between first and last are simply deleted. (from: TclHelp)

Examples:

 % lreplace  {a b c} 0 0 @
 @ b c
 % lreplace  {a b c} 1 1 
 a c

If the last argument of lreplace was made optional, one could essentially just alias ldelete to lreplace CmCc

TFW Feb 20, 2004 - Inplace replacements

Often time we have a long list we want to replace "in-place", that is without copying the list to a new altered list. Using the K combinator it is quite easy. (Note 8.5 is needed for the {expand} syntax)

 proc lipreplace {_list first last args} {
    upvar $_list list
    set list [lreplace [K $list [set list ""]] $first $last {expand}$args]
 }
 > set A [list 1 2 3]
 1 2 3
 > lipreplace A 1 end c d
 1 c d