MIPS: Calculating Displacement -

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working through problems , confused calculating displacement.

some examples are:

top:     addi    $s2, $s2, -1 addi    $s1, $s1, 1 bne $s2, $0, top

assume top has value of 0x1000 0008.

top:    bne     $s1, $s2, end addi    $s1, $s1, 1         end:    j   top

assume top has value of 0x1000 0008.

what displacement in bne instructions?

can explain how calculate these? thanks.

the absolute address doesn't matter. matters distance between jump , target.

to calculate offset, calculate distance between jump target , instruction following bne. in first example distance 12 bytes because there 3 instructions between target label , instruction following bne, , each instruction 4 bytes in size. , since backwards jump has negative offset, i.e. -12.

since instructions need word aligned (4 bytes), offset stored in instruction word shifted 2 bits right (since 2 least significant bits 00 anyway). arithmetic shift, meaning sign bit preserved. need take -12 , arithmetic right shift 2 bits actual offset gets stored in instruction word. gets easier if view -12 in hexadecimal form, 0xfff4. shifting 0xfff4 gives 0xfffd, we'd put in instruction word.

in second example offset positive, apart way calculate same.


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