gas - 'Wrong' usage of carry flag in ARM subtract instructions? -

the arm subtraction instructions carry (sbc, rsc) interpret carry flag (c) as:

• 0 means borrow
• 1 means no borrow

why carry flag c inversed make arithmetic?

sbc r0, r1, r2 @ r0 = r1 - r2 - !c 

we know grade school that

a = b - c = b + (-c) 

and elementary programming classes negate number in twos complement invert number , add 1

a = b + (~c) + 1 

and works out because when feed adder logic our values invert second operand , invert carry in. (there no subtract logic use adder subtraction)

so processor implementations choose invert carry out. inverting stuff subtract, , raw carry out of msbit on normal subtract (without borrow) 1, when there borrow carry out 0, inverting carry out can call "borrow" instead of carry (or instead of /borrow or !borrow or borrow_n)

which ever way it, if have subtract borrow, need either invert or not invert carry in depending design choices carry out on subtract (normal without borrow).

so subtraction is

a = b + (~c) + 1 

and when cascade subtraction (sbc) if there no borrow

a = b + (~c) + 1 

but if there borrow next level needs have 1 removed sbc becomes

a = b + (~c)  

the carry in 0 in case. adder logic perspective when subtract borrow naturally have invert second operand, carry in 0 if have borrow , carry in 1 if dont.

the arm docs on sbc say

rd = rd - rm - not(c flag)

so if need take 1 away due prior borrow c needed 0 else c 1 going in.

the subtractions say

c flag = not borrowfrom(operation)

so if need borrow borrow true c flag not true or 0. if dont need borrow, borrow false 0, , not borrow 1. matches up, if needed borrow going sbc c needs 0.

so arm not appear modify carry out adder. , not need invert carry in on sbc.