SUB- Subtract without Carry

Subtracts two registers and places the result in the destination register Rd.

Operation:

(i)Rd ← Rd - Rr

Syntax: Operands: Program Counter:

(i)SUB Rd,Rr 0 ≤ d ≤ 31, 0 ≤ r ≤ 31 PC ← PC + 1

16-bit Opcode:

0001

10rd

dddd

rrrr

I	T	H	S	V	N	Z	C
-	-	⇔	⇔	⇔	⇔	⇔	⇔

H: $\bar{Rd3} • Rr3 + Rr3 • R3 + R3 • \bar{Rd3}$

Set if there was a borrow from bit 3; cleared otherwise

S: N ⊕ V, For signed tests.

V: $Rd7 Rd7 • \bar{Rr7} • \bar{R7} + \bar{Rd7} • Rr7 • R7$

Set if two's complement overflow resulted from the operation; cleared otherwise.

N:R7

Set if MSB of the result is set; cleared otherwise.

Z: $\bar{R7} • \bar{R6} • \bar{R5} • \bar{R4}$ $• \bar{R3} • \bar{R2} • \bar{R1} • \bar{R0}$

Set if the result is $00; cleared otherwise.

C: $\bar{Rd7} • Rr7 + Rr7 • R7 + R7 • \bar{Rd7}$

Set if the absolute value of the contents of Rr is larger than the absolute value of Rd; cleared otherwise.

R (Result) equals Rd after the operation.

Example:

sub r13,r12 ; Subtract r12 from r13
brne noteq ; Branch if r12⇔r13
...
noteq: nop ; Branch destination (do nothing)

Words: 1 (2 bytes)

Cycles: 1

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