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in reply to Re^2: Array question

Perhaps I was less than careful with my use of the word "add". I simply meant you can do this:
```my @arr;
\$arr = "foo";

But it is actually pretty straightforward to insert elements into an array, shifting all later elements, using splice:

```my @a = ('foo','baz');
splice(@a,1,0,'bar');
print "@a\n";

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Re^4: Array question
by ww (Archbishop) on Jun 15, 2008 at 02:29 UTC
++, sgifford and many thanks!

And, note the clarity (and, in this case, the particularity and relevance) of sgifford's explanation, as contrasted with

```perldoc -f splice
splice ARRAY,OFFSET,LENGTH,LIST
splice ARRAY,OFFSET,LENGTH
splice ARRAY,OFFSET
splice ARRAY
Removes the elements designated by OFFSET and LENGTH from an
array, and replaces them with the elements of LIST, if any. I
+n
list context, returns the elements removed from the array. In
scalar context, returns the last element removed, or "undef"
+if
no elements are removed. The array grows or shrinks as
necessary. If OFFSET is negative then it starts that far from
the end of the array. If LENGTH is omitted, removes everythin
+g
from OFFSET onward. If LENGTH is negative, removes the elemen
+ts
from OFFSET onward except for -LENGTH elements at the end of
+the
array. If both OFFSET and LENGTH are omitted, removes
everything. If OFFSET is past the end of the array, perl issu
+es
a warning, and splices at the end of the array.

The following equivalences hold (assuming "\$[ == 0 and \$#a >=
\$i" )

push(@a,\$x,\$y)      splice(@a,@a,0,\$x,\$y)
pop(@a)             splice(@a,-1)
shift(@a)           splice(@a,0,1)
unshift(@a,\$x,\$y)   splice(@a,0,0,\$x,\$y)
\$a[\$i] = \$y         splice(@a,\$i,1,\$y)

Example, assuming array lengths are passed before arrays:

sub aeq {   # compare two list values
my(@a) = splice(@_,0,shift);
my(@b) = splice(@_,0,shift);
return 0 unless @a == @b;       # same len?
while (@a) {
return 0 if pop(@a) ne pop(@b);
}
return 1;
}
if (&aeq(\$len,@foo[1..\$len],0+@bar,@bar)) { ... }