Lesson 21 · 8XY5 and 8XY7 - subtract with borrow

8XY5 and 8XY7 - subtract with borrow

Subtraction reports a borrow, but with an inverted flag - VF is 1 when there is NO borrow. Today you implement 8XY5 (VX minus VY) and 8XY7 (VY minus VX), pinning that inverted flag.

The goal

Implement 8XY5 and 8XY7 so they subtract with an 8-bit result and set VF to 1 when there is no borrow.

Start here - the target

TO DO

Scenario: The subtracts set VF to the NOT-borrow flag

GivenV0 holds 0x05 and V1 holds 0x03, about to execute 0x8015 (V0 = V0 - V1)

WhenStep runs it

ThenV0 becomes 0x02 and VF becomes 1 (no borrow, because V0 >= V1)

Andfrom V0 = 0x03, V1 = 0x05, 0x8015 gives V0 = 0xFE (wrapped) and VF = 0 (a borrow occurred); and 0x8017 (V1 - V0) gives V0 = 0x02 with VF = 1

Background

Subtraction mirrors the add, with one twist that trips everyone up: the flag is inverted. For 8XY5 (VX = VX - VY), VF is set to 1 when there is no borrow - that is, when VX is greater than or equal to VY - and 0 when the subtraction would go negative and wraps around. So 0x05 - 0x03 stores 0x02 with VF = 1, while 0x03 - 0x05 wraps to 0xFE with VF = 0. Think of VF here as “the result is valid / did not underflow,” which is the opposite polarity of the carry flag.

8XY7 is the same operation with the operands reversed: VX = VY - VX, with VF = 1 when VY >= VX. It exists so a program can subtract in either direction without first copying registers around. Both keep only the low byte of the result and, like 8XY4, write VF after the store. Pin all three facets - a clean subtract, one that borrows, and the reversed form - so the inverted flag is verified in both directions.

Make it work

case 0x5: // 8XY5: VX = VX - VY
  // subtract in a wide type, keep the low byte in VX, then set VF
  // to 1 when the minuend was >= the subtrahend (no borrow), else 0.
case 0x7: // 8XY7: VX = VY - VX  (same rule, operands swapped)

CheckpointDONE

8XY5 and 8XY7 subtract with wraparound and an inverted borrow flag. Commit and stop here.