Clarification of Question by
motts786-ga
on
04 May 2005 15:46 PDT
clarification on 'A' Q and Q_1:
the multiplier and multiplicand are placed in the Q and
M registers, respectively. There is also a 1-bit register placed
logically to the right of the
least significant bit (Q0) of the Q register and designated Q-1; its
use is explained shortly.
The results of the multiplication will appear in the A and Q
registers. A and Q-1 are
initialized to 0. As before, control logic scans the bits of the
multiplier one at a time.
Now, as each bit is examined, the bit to its right is also examined.
If the two bits are the
same (1-1 or 0-0), then all of the bits of the A, Q, and Q-1 registers
are shifted to the right
1 bit. If the two bits differ, then the multiplicand is added to or
subtracted from the A
register, depending on whether the two bits are 0-1 or 1-0. Following
the addition or
subtraction, the right shift occurs. In either case, the right shift
is such that the leftmost bit
of A, namely An-1, not only is shifted into An-2, but also remains in
An-1. This is required
to preserve the sign of the number in A and Q. It is known as an
arithmetic shift, because
it preserves the sign bit.