1.2. CHAPTER 2: REPRESENTING AND MANIPULATING INFORMATION
3
There are many solutions to this problem, but it is a little bit tricky to write one that works for any word
size. Here is our solution:
code/data/shift-ans.c
1 int int_shifts_are_arithmetic()
2 {
3 int x = ˜0; /* All 1’s */
4
5 return (x >> 1) == x;
6 }
code/data/shift-ans.c
The above code peforms a right shift of a word in which all bits are set to 1. If the shift is arithmetic, the
resulting word will still have all bits set to 1.
Problem 2.45 Solution:
This problem illustrates some of the challenges of writing portable code. The fact that 1<<32 yields 0 on
some 32-bit machines and 1 on others is common source of bugs.
A. The C standard does not define the effect of a shift by 32 of a 32-bit datum. On the SPARC (and
many other machines), the expression x << k shifts by
, i.e., it ignores all but the least
significant 5 bits of the shift amount. Thus, the expression 1 << 32 yields 1.
B. Compute beyond_msb as 2 << 31.
C. We cannot shift by more than 15 bits at a time, but we can compose multiple shifts to get the
desired effect. Thus, we can compute set_msb as 2 << 15 << 15, and beyond_msb as
set_msb << 1.
Problem 2.46 Solution:
This problem highlights the difference between zero extension and sign extension. It also provides an excuse
to show an interesting trick that compilers often use to use shifting to perform masking and sign extension.
A. The function does not perform any sign extension. For example, if we attempt to extract byte 0 from
word 0xFF, we will get 255, rather than
.
B. The following code uses a well-known trick for using shifts to isolate a particular range of bits and to
perform sign extension at the same time. First, we perform a left shift so that the most significant bit
of the desired byte is at bit position 31. Then we right shift by 24, moving the byte into the proper
position and peforming sign extension at the same time.
code/data/xbyte.c
1 int xbyte(packed_t word, int bytenum)
2 {
