18-JoP-June-08
1 pages
Published by
sgganesh
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All rights reserved
102 june 2008 | LInuX For You | www.
openITis.
com
T
he following trick using three consecutive ex-or
operations for swapping two variables without
using a temporary is well known and particularly
popular among students:
i ^= j; j ^=...
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102 june 2008 | LInuX For You | www.
openITis.
com
T
he following trick using three consecutive ex-or
operations for swapping two variables without
using a temporary is well known and particularly
popular among students:
i ^= j; j ^= i; i ^= j;
Here, i and j are integer variables and it works well
(we covered a few pitfalls earlier with this).
For further
optimisation, one often sees this solution combining the
three statements into a single statement:
i ^= (j ^= (i ^= j));
Try this with your favourite C compiler; it usually
works.
This trick works based on the following
assumption: the values of i and j are modified in the RHS
(right-hand side) of the expression and the modified
values are expected to be used in the LHS (left-hand side)
of the expression.
However, this solution need not work and the updated
values of i and j need not get reflected when read
again.
This is because the compiler is free to optimise
the sequence of reads and writes, using temporaries
internally.
This problem is covered by an intricate and
difficult to understand technical issue known as ‘sequence
points’ (see en.
wikipedia.
org/wiki/Sequence_point).
For
example, if the initial value of i is 0, after executing the
statement i=i++, i might be 0 or 1: it is implementation
defined behaviour.
For the same reason, the expression i
^= (j ^= (i ^= j)) need not swap the two variables i and j
correctly—it depends on the compiler as to what we’ll get
as values for i and j.
We’ll take a different approach in this column
and cover Java to understand the implications of
such statements.
Unlike C, Java does not have code
segments leading to implementation defined behaviour
(i.
e.
, the programs written in Java will behave the
same way, irrespective of the Java compiler used).
For
the swap in a single statement case, let us see what
happens in Java:
class Swap {
S.
G.
GaneSh
In May and June last year, we covered a few traps and pitfalls in trying to swap two variables without
using a temporary.
This month, we’ll look at swapping variables without a temporary in a single
statement, and see how C and Java compilers treat them differently.
The Joy of
Programming
Traps and Pitfalls: Swapping Two Variables—Part III
S.
G.
Ganesh is a research engineer in Siemens (Corporate
Technology), Bangalore.
His latest book is “60 Tips on
Object Oriented Programming” (ISBN 978-0-07-065670-3),
published by Tata McGraw-Hill, New Delhi.
You can reach
him at sgganesh@gmail.
com
public static void main(String []s) {
int i = 3, j = 6;
System.
out.
println(“Before swap: i = “ + i + “ j = “ + j);
i ^= (j ^= (i ^= j));
System.
out.
println(“After swap: i = “ + i + “ j = “ + j);
}
}
The swap trick does not work! What happened? The
Java specification (section 15.
26.
2) says: “The value of
the left-hand side of a compound assignment is saved
before the right-hand side is evaluated”.
In other words,
the value of i and j are remembered and used for the
LHS values of i and j; so we do not get the values of i and
j swapped correctly.
To illustrate this, let us try the following: Precompute the result of (i ^ j) and remember it in a
temporary.
Now use that temporary in the LHS instead
of i; you’ll get the expected (correct) output of the
swapped values of two variables:
class SwapTest {
public static void main(String []s) {
int i = 3, j = 6;
int cache = (i ^ j);
System.
out.
println(“Before swap: i = “ + i + “ j = “ + j);
cache ^= (j ^= (i ^= j));
i = cache;
System.
out.
println(“After swap: i = “ + i + “ j = “ + j);
}
}
It works, right?
Well, don’t worry if you feel that the explanation for the
Java programs is not clear.
Next month, we’ll dig deeper by
looking at the byte codes generated for the swap code and
understand how it works.
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