Question

In a computer instruction format, the instruction length is 12 bits and the size of an address field is 5 bits. The system architect has already designed three 2-address instructions and thirty one 1-address instructions. How many 0-address instructions can still be possibly accommodated?

Answer 1

Total Number of possible Instructions= 2¹²=4096

size of address field = 5 bits.

There are 3 2-address Instruction, these 10 bits are used for address & Remaining 2 bits are used to select one out of the 3 possible Instructions.

Total bit pattern needed= 3* 2⁵*2⁵

                                             =3*2¹⁰

                                             =3*1024

                                              =3072

For thirty-one 1-address instruction, bit Patton needed=31*2⁵

^{=31*2*2*2*2*2}

=31*32

=992
Total number of used bit patterns = 3072(3 2-address)+992(thirty one 1-address)

=3072+992

=4064

Number of 0-address instructions possible=4096( Total Number of possible Instructions)-4064( Total number of used bit patterns)

=4096-4064

=32