我感到麻烦的是把这一组装成法典。 头1号车的价值是什么样的,我会怎样得到?
3242ba66 f6454118 movw r1, 0x5c18
3242ba6a 466f mov r7, sp
3242ba6c f6c0415a movt r1, 0xc5a
3242ba70 f2460002 movw r0, 0x6002
3242ba74 f6c0405a movt r0, 0xc5a
3242ba78 4479 add r1, pc
3242ba7a 4478 add r0, pc
3242ba7c 6809 ldr r1, [r1, #0]
<代码>movw后加<>movt,是将32-bit数值输入登记册的一种共同办法。 这相当于两种直接数值的OR-,movt为16上限。 在这种情况下,r1 = (去除现值和带;< 16) >(立即值)。
3242ba66 f6454118 movw r1, 0x5c18 // r1 = 0x5c18
3242ba6a 466f mov r7, sp
3242ba6c f6c0415a movt r1, 0xc5a // r1 = (r1 & 0xffff) | (0xc5a << 16)
3242ba70 f2460002 movw r0, 0x6002
3242ba74 f6c0405a movt r0, 0xc5a
3242ba78 4479 add r1, pc // r1 = r1 + pc
3242ba7a 4478 add r0, pc
3242ba7c 6809 ldr r1, [r1, #0] // r1 = *(r1 + 0)
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