ARMv7-M汇编：寄存器中未对齐的字节替换？

Question

Suppose we have three destination registers, r5, r6, and r7, and consider them to be an array of 12 bytes:

r5={a, b, c, d} r6={e, f, g, h} r7={i, j, k, l}

What I'd like to do is read eight bytes {0, 1, 2, 3, 4, 5, 6, 7} from a given memory location, and copy them into the three destination registers to give

r5={a, 0, 1, 2} r6={3, 4, 5, 6} r7={7, j, k, l}

In other words, I'd like to replace bytes 1-8 of of the array in r5-r7 with the eight bytes read from ROM.

I can do this in six instructions:

ldrd    r0, r1, [r3]
bfi     r5, r0, #8, #24
ubfx    r6, r0, #24, #8
bfi     r6, r1, #8, #24
bfc     r7, #0, #8
orr     r7, r7, r1, lsr #24

Can it be done in fewer? Assume the absence of floating-point unit.

(Background: this is part of a blitting routine, several pixel patterns are read from memory and composited into a set of registers which is then written to a display buffer with stmia .)

Answer 1

I'm hesitant to post the following, because it doesn't solve the problem as posed - it assembles the output in memory rather than in the registers.

But for what it's worth:

r5,r6,r7 - abcdefghijkl      (as per the question)
r3       - input array ptr   (as per the question)
r2       - output array ptr

                        r5   r6   r7    [r3]         r0    r1        [r2]
                        abcd efgh ijkl  01234567
ldrd    r0, r1, [r3]                                 0123  4567      ------------
str     r5, [r2, #0]                                                 abcd-------- 
str     r7, [r2, #8]                                                 abcd----ijkl 
str.w   r0, [r2, #1]                                                 a0123---ijkl 
str.w   r1, [r2, #5]                                                 a01234567jkl

Note that:

• The three output words are assembled in four writes to memory, two of which are unaligned.
• The two str instructions are 16-bit encodings.