107 lines
4.7 KiB
Python
107 lines
4.7 KiB
Python
# -*- coding: utf-8 -*-
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"""独立校验交付的 .face（只看交付物 + 固件镜像，不看中间产物）。
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rome: replace -> dist/S4Chg.face pay 应等于**改后镜像**的字节，fp 应等于**原厂**前 64 B
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restore -> dist/S4ChgRestore.face pay 应等于**原厂**字节， fp 应等于**改后镜像**前 64 B
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① 从 .face 里抠出 Lua 表（pay / rb），按表里的偏移取字节
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② 每条 pay 长度必须 == 原厂槽位长度（等长覆盖的底线），且能按 LVGL RLE 解出 464x464
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③ rb 每条必须是 64 B 指纹，且等于另一侧字节的前 64 B（版本校验数据正确）
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④ 目标镜像的 inode 表 / superblock / 分区长度与原厂逐位一致
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"""
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import os, re, sys
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sys.path.insert(0, os.path.dirname(os.path.abspath(__file__)))
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from paths import PROJ, INPUTS, WORK
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from romfs import Romfs
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import lvgl_bin as L
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ROLE = (sys.argv[1] if len(sys.argv) > 1 else 'replace').lower()
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NAME = 'S4Chg' if ROLE == 'replace' else 'S4ChgRestore'
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FACE = os.path.join(PROJ, 'dist', NAME + '.face')
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STOCK = os.path.join(INPUTS, 'target', 'vela_system.bin')
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PATCHED = os.path.join(WORK, 'charge', 'vela_system_charge.bin')
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DIRP = 'charging/'
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FP_LEN = 64
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ok = fail = 0
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def ck(cond, msg):
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global ok, fail
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if cond:
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ok += 1
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else:
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fail += 1
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print(' !! ' + msg)
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face = open(FACE, 'rb').read()
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src = face.decode('latin1')
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m = re.search(r'pay = \{(.*?)\n \},', src, re.S)
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m2 = re.search(r'rb = \{(.*?)\n \},', src, re.S)
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ck(m is not None and m2 is not None, '在 .face 里找不到 pay / rb 表')
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pay = [tuple(int(x) for x in r) for r in re.findall(r'\{(\d+),(\d+),(\d+)\}', m.group(1))]
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rb = [tuple(int(x) for x in r) for r in re.findall(r'\{(\d+),(\d+),(\d+)\}', m2.group(1))]
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print('%s：pay %d 条，rb %d 条' % (NAME, len(pay), len(rb)))
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stock_b = open(STOCK, 'rb').read()
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patch_b = open(PATCHED, 'rb').read()
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c = re.search(r"local ROLE = '(\w+)'", src)
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ck(c is not None and c.group(1) == ROLE, 'Lua 里的 ROLE 不是 %s' % ROLE)
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EXPECT_PAY = patch_b if ROLE == 'replace' else stock_b
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EXPECT_FP = stock_b if ROLE == 'replace' else patch_b
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stock = Romfs(STOCK)
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slots = {}
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for p, i in stock.entries():
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if i.type == 1 or not p.startswith(DIRP) or not p.endswith('.bin') or i.size == 0:
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continue
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slots[i.data_off] = (p, i.size)
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canary = [r for r in pay if r[1] == 0 and r[2] == 32]
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ck(len(canary) <= 1, 'canary 记录不止一条')
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if canary:
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ck(face[canary[0][0]:canary[0][0] + 32] == stock_b[:32], 'canary 不对')
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data = [r for r in pay if not (r[1] == 0 and r[2] == 32)]
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bad_len = bad_dec = bad_bytes = 0
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for o, t, ln in data:
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blob = face[o:o + ln]
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if t not in slots or slots[t][1] != ln:
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bad_len += 1; continue
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if EXPECT_PAY[t:t + ln] != blob:
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bad_bytes += 1
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h = L.parse_header(blob)
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if h is None or not (h['flags'] & L.FLAG_COMPRESSED) or h['w'] != 464 or h['h'] != 464:
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bad_dec += 1; continue
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# ★ 设备端解码器要求流解出**正好 usize 字节**（1024 + w*h + 1），少一字节就整帧不画
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import struct as _s
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_m, _cs, _us = _s.unpack('<III', blob[12:24])
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if len(L.rle_decompress(blob[24:24 + _cs], 1, None)) != _us:
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bad_dec += 1; continue
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try:
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if L.decode_lvgl_bin(blob)[1] is None:
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bad_dec += 1
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except Exception:
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bad_dec += 1
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ck(bad_len == 0, '有 %d 条长度 != 原厂槽位长度' % bad_len)
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ck(bad_dec == 0, '有 %d 条解不出 464x464' % bad_dec)
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ck(bad_bytes == 0, '有 %d 条 != %s 的字节' % (bad_bytes, '改后镜像' if ROLE == 'replace' else '原厂镜像'))
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print(' pay 检查 %d 条：长度全对 %s，可解码 %s，字节与%s一致 %s'
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% (len(data), bad_len == 0, bad_dec == 0, '改后镜像' if ROLE == 'replace' else '原厂镜像', bad_bytes == 0))
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bad_fp = 0
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for o, t, ln in rb:
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if ln == 0 or (t == 0 and ln == 32): # canary 那条的指纹是空表
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continue
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if ln != FP_LEN or t not in slots or face[o:o + ln] != EXPECT_FP[t:t + ln]:
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bad_fp += 1
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ck(bad_fp == 0, '有 %d 条指纹不对（应为另一侧前 64 B）' % bad_fp)
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print(' rb 指纹 %d 条：全部 64 B 且与另一侧一致 %s' % (len(rb), bad_fp == 0))
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p2 = Romfs(PATCHED)
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a = [(p, i.off, i.next, i.spec, i.size, i.cksum) for p, i in stock.entries()]
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b = [(p, i.off, i.next, i.spec, i.size, i.cksum) for p, i in p2.entries()]
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ck(a == b, 'inode 表变了')
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ck(stock.sb_size == p2.sb_size and stock.sb_cksum == p2.sb_cksum, 'superblock 变了')
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ck(len(stock_b) == len(patch_b), '分区长度变了')
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ck(len(face) < 20 * 1048576, '交付物体积异常：%d B' % len(face))
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print('\n%s —— %d 项通过，%d 项失败' % ('全部通过' if fail == 0 else '有失败', ok, fail))
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sys.exit(1 if fail else 0)
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