Question

please give full explanation.

Suppose the program counter (PC) is set to 0x1000FFFF. Is it possible to
use the jump ( j ) MIPS assembly instruction to jump directly to the instruction at 0x10003000?
If yes, write the corresponding assembly instruction(s). If not, explain why and give other MIPS
instruction(s) that can jump to this target address. (For this problem, assume that the instruction
memory starts from 0x00000000 and goes to 0xFFFFFFFF.)

Answer 1

The Program counter is set to 0x1000FFFF.

The jump (j) instruction is a 32-bit instruction. It uses 6 bits for “opcode” and the remaining (32 - 6) = 26 bits for the “target address”.

While executing the jump statement, the 26-bit address changes into a 32-bit address by shifting left by two bits and taking the upper 4 bits of the program counter.

The program counter is set to 0x1000FFFF. The binary equivalent is

0000 0000 0001 0000 0000 0000 1111 1111 1111 1111.

The highest possible value of jump with the largest value in the lower 26 bits is 0x13FFFFFF.

The 26 bits are shifted left by two bits to form the 28-bit value 0x1FFFFFFC.

The upper order four bits of the program counter, that is 0x2 are appended to this 28-bit value to form the 32-bit value, that is 0x2FFFFFFFC.

As the address 0x2FFFFFFFC is the maximum size of jump, it is not possible to jump (j) in MIPS to the address 0x10003000.

The jump instruction cannot modify the upper four bits.

So, it is impossible to set the PC to the address as 0x10003000.

Hope this helps.