FA-89634 / Instruction set emulation / Member archive

Decimal carry derived from 8-bit overflow of the adjust · case 04

A high-digit adjust that stays below 0x100 loses the decimal carry.

Member previewVariant 4 · 3 implementations · 10 checks per implementation

Case contract

Input [xs, ys, c]: little-endian packed-BCD byte lists and a carry-in. For each byte: s = a+b+c; AF = low-nibble sum (with carry) > 0x0F; CF = s > 0xFF; al = s & 0xFF. Decimal adjust: if (al & 0x0F) > 9 or AF then al += 6 (8-bit wrap); then if the pre-adjust al > 0x99 or CF then al += 0x60 (8-bit wrap) and carry-out = 1, else carry-out = 0. The carry-out feeds the next byte. Return [result bytes, final carry].

Why this case matters

Decimal-adjust instructions are among the hardest to emulate; the flag inputs and the pre-adjust comparison decide the result.

One recorded failure

Sample boundary fixture

This sample comes from the broken implementation of a controlled reproducer.

Boundary fixture	Actual	Expected	Outcome
binary carry from high digits	`[[152, 3], 1]`	`[[152, 4], 1]`	Failed

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