NAME

       lreplace - Replace elements in a list with new elements

SYNOPSIS

       lreplace list first last ?element element ...?
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DESCRIPTION

       lreplace  returns  a  new  list formed by replacing zero or more elements of list with the
       element arguments.  first and last are index values specifying the first and last elements
       of  the  range  to  replace.   The index values first and last are interpreted the same as
       index values for the command string index, supporting simple index arithmetic and  indices
       relative  to  the  end  of  the  list.  0 refers to the first element of the list, and end
       refers to the last element of the list.

       If either first or last is less than zero, it is considered to refer to before  the  first
       element of the list. This allows lreplace to prepend elements to list.  If either first or │
       last indicates a position greater than the index of the last element of the  list,  it  is │
       treated  as  if  it is an index one greater than the last element. This allows lreplace to │
       append elements to list.

       If last is less than first, then any specified elements will be  inserted  into  the  list
       before the element specified by first with no elements being deleted.

       The  element  arguments specify zero or more new elements 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.

EXAMPLES

       Replacing an element of a list with another:

              % lreplace {a b c d e} 1 1 foo
              a foo c d e

       Replacing two elements of a list with three:

              % lreplace {a b c d e} 1 2 three more elements
              a three more elements d e

       Deleting the last element from a list in a variable:

              % set var {a b c d e}
              a b c d e
              % set var [lreplace $var end end]
              a b c d

       A procedure to delete a given element from a list:

              proc lremove {listVariable value} {
                  upvar 1 $listVariable var
                  set idx [lsearch -exact $var $value]
                  set var [lreplace $var $idx $idx]
              }

       Appending elements to the list; note that end+2 will initially be treated as if  it  is  6 │
       here,  but both that and 12345 are greater than the index of the final item so they behave │
       identically:                                                                               │

              % set var {a b c d e}                                                               │
              a b c d e                                                                           │
              % set var [lreplace $var 12345 end+2 f g h i]                                       │
              a b c d e f g h i                                                                   │

SEE ALSO

       list(3tcl),  lappend(3tcl),  lindex(3tcl),  linsert(3tcl),  llength(3tcl),  lsearch(3tcl),
       lset(3tcl), lrange(3tcl), lsort(3tcl), string(3tcl)

KEYWORDS

       element, list, replace