Question

The machine language form of a jump instruction (j) is 00001001001100101000110001010110. The address of the jump instruction (PC) is 0x20201018. What address this the program will jump to?

Answer 1

The answer for the above mentioned question is explained below ::

The program will jump to the following address :

00100100110010100011000101011000

Explanation is as follows ::

Given that , the machine language form for jump instruction is

000010 01001100101000110001010110

where , first six bits represent opcode i.e., here 000010 represents the opcode j .

The remaining 26 bits represents the address , to get the final address to which the given instruction jumps , we have to multiply the 26 bit address field by 4 , so after multiplication of 26 bits by 4 we get 28 bits and we append most significant 4 bits from the PC , i.e., here PC is 0x20201018 here most significant 4 bits is 0010 . Therefore the above program will jump to the address 00100100110010100011000101011000 .

Explanation in terms of Addresses ::

Machine language form for jump instruction is ::

00001001001100101000110001010110

Here first 6 bits represents opcode j i.e., 000010 represents j and remaining 26 bits represents the target address i.e., 01001100101000110001010110

To get the resultant address to which the given program jump is first 4 bits are taken from the most significant 4 bits of PC and next 28 bits are taken from the (26 bit *4) .

i.e., Here most significant 4 bits of PC is 0010 .

and 26 bits *4 is 0100110010100011000101011000

Therefore the given program jump to the address

00100100110010100011000101011000