Can someone explain why this following assembly line makes sense ?
beq $0, $0, 1
Note that $0 refers to a register that always have the value of 0. So we are saying if $0 = $0 then go to PC + 4 + 1 else go to next instruction.
My confusion comes from the immediate field of beq instruction which is 1. Does this mean we are going to address PC + 5 ??? Doesn't MIPS requires alignment when accessing memory and all memory locations have to be divisible by 4 ?
Note that the book says that this instruction just skips the next instruction.
The semantic of beq $t, $s, offset is
if ($t == $s)
PC = PC + 4 + 4 * offset;
else
PC = PC + 4
Simply put the PC is always advanced by 4 by the time the instruction is executed and the immediate is assumed missing the lower two bits since they are always zero and can be reintroduced with a shift (offset * 4 = offset << 2).
beq $0, $0, 1 just skip the next instruction as it tautologically set PC = PC + 8.