This document discusses bit manipulation techniques in C programming. It presents 6 questions with explanations. Q1 checks if a number is a power of 2. Q2 returns the lowest bit set in a number. Q3 expresses the 1's complement of a number using bitwise operators. Q4 returns the maximum of two numbers. Q5 expresses the addition of two numbers using bitwise operators. Q6 resets a number to another number using bitwise XOR. The document provides concise yet detailed explanations of how each bit manipulation expression works at the bit level. It discusses concepts like 1's complement, lowest bit, power of 2, and using bitwise logic to find maximum and express addition.

1. The Joy of
Programming
How ‘C’mart Are You?
Some Bit-manipulation S.G. GANESH
Tricks and Techniques
In this column, lets see few interesting bit-manipulation tips; such techniques are useful and
widely used in advanced low-level programming tasks such as writing software emulators
and device drivers.
It is assumed that the underlying implementation follows EXORing i with all 1s results in resetting the bits
2’s complement representation for integers. that are set; so it’s the same as ~i!
Assume that i, j and r are integers. In ~i = -(i + 1) implies ~i = (-i -1). Move -1 to the left
Q1: What does the expression “(i & (i – 1))” do? side and the expression is: (~i + 1 = -i); now you can
A: It checks if i is a power of 2 or not! read it as “1’s complement of i plus 1 is 2’s
GUEST COLUMN
Q2: What does the expression (i & (-i)) do? complement of i”, which is correct!
A: It returns the lowest bit set in an integer (which is a Q 4: (j + ((i - j) & -(i < j)) – let’s read it as: With
power of 2)! j (it can be i also), add the difference
Q3: How can you express ~i in terms of other (preferably between the two, and (re)set the sign bit
bit-wise) operators? for the value of the difference between the
A: (i ^ (~0)) and –(i + 1). two. If j is less than i, the expression adds
Q4: What does the expression (j + ((i - j) & -(i < j)) do? the difference between the two; or else it
A: This code results in the maximum of two integers i subtracts the difference between the two.
and j. Following the same logic (j - ((i - j) & -(i <
Q5: How can you express the expression (i + j) in terms j)) is the minimum of i and j.
of bit-wise ‘&’ and ‘|’ operations? Q 5: (i | j) sets the bits that are only set in either
A: ((i & j) + (i | j)) i or j. (i & j) shows the value that’s set in
Q6: What is the significance of the expression r = (i ^ j ^ both the values. Adding these two results
r)? has the effect of retaining the bits set, the
A: If r is equal to i, r will be reset to j; else if r is equal to bits set. Then adding the bits set in both the
j, r will be reset to i! values again with the result, if you closely
observe, is the same as, it is the same as
That’s enough; let’s see the explanation. (i + j).
Q1. When i is a value that’s a power of 2, only a single bit Q 6: If r is equal to i, it will nullify the effect of it
will be set in that integer. Now, when you subtract 1 in the expression (i ^ j ^ r) resulting in j;
from that value, that bit will be reset and will result the same happens if r is equal to j. (This is the same
in setting the lower bits. For (i & (i – 1)), when it is a trick used to re-write a doubly linked list—with
power of 2, the resulting two bit patterns will have previous and next pointers—with a single pointer,
no common bits set, so the expression becomes false which has the EXORed result of the previous and
(in that case it’s a power of 2); else it isn’t. next pointers!)
Q 2: -i is 1’s complement of i plus 1; when you invert all
the bits in an integer and add 1 to it, all the set bits S.G. Ganesh is an engineer in Hewlett-Packard’s C++
in the end are reset to zero. When we do ‘&’ of these compiler team. He has authored a book “Deep C” (ISBN 81-
two values, we get only the lowest bit set (since all 7656-501-6). He is also a member of the ANSI/ISO C++
other bits will have complementary values set in Standardisation committee (JTC1/SC22/WG21),
representing HP. You can reach him at
them).
sgganesh@gmail.com.
Q 3: ~i is 1’s complement of i; since ~0 is all 1’s set and
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