SPOILER Re: SPUG: scalar swap challenge

[Lee Wilson]

Sheesh, I sent this earlier, but I think I sent it to an individual rather 
than the list.  Doh =)

One of my favorite interview questions is to ask what this code does.  
Assume A and B are of int type.

B=(A/2+A/2+(A%2==1?1:0))+(A=B)-B;

This line can reduce to:
B = A + (A=B) - B;

Answer: it swaps A and B without using a temp var, and in one line of 
code.  I think this second line would work in perl for any scalar type, 
actually (assuming you put $A and $B =) ).



On Wed, 8 Oct 2003, Tom Legrady wrote:

>     # Your $a/$b swap code goes here.
>     $a ^= $b;
>     $b ^= $a;
>     $a ^= $b;


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