Question

The data path shown in the figure computes the number of 1s in the 32-bit input word corresponding to an unsigned even integer stored in the shift register. The unsigned counter, initially zero, is incremented if the most significant bit of the shift register is 1.


The microprogram for the control is shown in the table below with missing control words for microinstructions $I_1,I_2,.....,I_n$

Micro- instruction	reset_counter	shift_left	load_output
BEGIN	1	0	0
$I_1$	?	?	?
:	:	:	:
$I_n$	?	?	?
END	0	0	1

The counter width (k), the number of missing microinstructions (n), and the control word for microinstructions $I_1,I_2,.....,I_n$ are, respectively.

• A. 5, 31, 010
• B. 5, 32, 010
• C. 32, 5, 010
• D. 5, 31, 011

Answer 1

The correct option is A 5, 31, 010

Since the number is even number i.e. LSB bit must be 0, so maximum 31 1's presents, so loop will be runs for 31 times i.e. number of missing instruction are 31.

The input word is of 32 bits long, so width of counter = log(32) = 5 bits.

To count number of 1's, reset counter will be reset i.e, 0. Shift left = 1 i.e., decrease number of 1's by 1 in each iteration and load output is 0 for all instructions except "END".