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There are two objects:

Object1, Object2

You can easily swap them by:

Object3 = Object1
Object1 = Object2
Object2 = Object3

However, you must solve this problem by using no temporary object and Object1 should be changed to Object 2. (and vice versa)

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closed as not a real question by Peter Taylor, Steven Rumbalski, boothby, vsz, grc Nov 13 '12 at 7:07

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

Here we have a prime example of what is wrong with challenges that don't have "an objective winning criteria" as currently required by the FAQ: three different answer that are all completely correct. So how will people judge them? All that we have here is a popularity context. – dmckee Nov 11 '12 at 20:31
@dmckee, the problem here isn't lack of objective winning criteria, but of any winning criteria. – ugoren Nov 11 '12 at 21:18


I'll say it first.

object1, object2 = object2, object1
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Also possible in Go. However, I think a temporary variable may still be made in the compiled code. – Max Morin Nov 12 '12 at 15:22
This is also valid Ruby. – steenslag Nov 15 '12 at 20:16


A standard feature:

a = 1; b = 2;
Print[a, " ", b];
{a, b} = {b, a};
Print[a, " ", b];


1 2

2 1
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In PostScript, it's common to use the stack instead of named variables, so a simple


would do the job. But if you have variables, you can do it like

/a 1 def
/b 2 def
/a b /b a def def
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If you have access to the memory address of these objects, you can swap them by xor-ing their addresses.

// return 0 for success, 1 for error
int swap(void *object1, void *object2, int size)
    for (int i = 0; i < size; i++)
        char *x = (char *)(object1 + i);
        char *y = (char *)(object2 + i);
        if (x == y)
            return 1;
        *x ^= *y;
        *y ^= *x;
        *x ^= *y;
    return 0;

per dmckee suggestion for clarity

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The solution is to pass a size as well, cast to char* and then xor swap the characters guarding against the same-value failure mode. Not as pretty, but standards compliant and working. – dmckee Nov 11 '12 at 21:44
Only answer which doesn't do a,b = b,a. +1 :) – beary605 Nov 12 '12 at 22:16


($object1, $object2) = ($object2, $object1);

Using Perl's list assertions feels like cheating, but it's a pretty common feature.

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Objects / arrays / resources:

list($a, $b) = [$b, $a];

Integers / booleans / strings of equal length:

$a ^= $b ^= $a ^= $b;
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As i have to include more than 30 characters, here's a link to a Forth primer

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