plucky (3) lreplace.3tcl.gz

Provided by: tcl9.0-doc_9.0.1+dfsg-1_all bug

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),  lassign(3tcl),  ledit(3tcl),  lindex(3tcl),  linsert(3tcl),   llength(3tcl),
       lmap(3tcl),   lpop(3tcl),  lrange(3tcl),  lremove(3tcl),  lrepeat(3tcl),  lreverse(3tcl),  lsearch(3tcl),
       lseq(3tcl), lset(3tcl), lsort(3tcl), string(3tcl)

KEYWORDS

       element, list, replace