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Programming-Idioms

  • Ruby
  • Perl

Idiom #322 replace value of variable with new one and return old value

use strict;

my $old;
my $new = 'new thing';
my $x = 'old thing';

($old,$x) = ($x, $new);

When use strict is in effect, variables must be predeclared.
no strict 'vars';
($old,$x) = ($x, $new);

This works in the default mode of no strict 'vars' where variables need not be pre-declared. See the other implementation for the more normal case where the use strict pragma is in effect.
old, x = x, new
utility
std::exchange(x, v);
New implementation...
< > ⌨
pi 
use strict;

my $old;
my $new = 'new thing';
my $x = 'old thing';

($old,$x) = ($x, $new);
no strict 'vars';
($old,$x) = ($x, $new);
old, x = x, new
std::exchange(x, v);
old, x = x, new
generic function replace<T>(var X: T; NewValue: T):T;
begin
  Result := X;
  X := NewValue;
end; 

begin
  x := someoldvalue;
  y := specialize replace<TSomeType>(x, somenewvalue);
end.
old, x = x, new
std::mem::replace(&mut x, v);