Question

A 2-Byte long machine insturction is stored in memory at address $(710{)}_{16}$. The branch is made to location $(630{)}_{16}$ . What is the signed offset value (in bytes) present in the instruction?

Answer 1

option (d)

In PC- relative mode:

Target address = PC value + offset

When current instruction will be in calculating target address phase.

then PC- value will be address of next instruction:

PC = 710 + 2 = $(712{)}_{16}$

Target address = PC value + offset

$(630{)}_{16}=(712{)}_{16}+offset$

offset = $-(E2{)}_{16}$