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Re: memcpy to an unaligned address

On Aug 2, 2005, at 10:32 AM, Shaun Jackman wrote:
In a typical Ethernet/IP ARP header the source IP address is
unaligned. Instead of using...
    out->srcIPAddr = in->dstIPAddr;
... I used...
    memcpy(&out->srcIPAddr, &in->dstIPAddr, sizeof(uint32_t));
... to account for the unaligned destination. This worked until gcc 4,
which now generates a simple load/store.
    ldr     r3, [r6, #24]
    adds    r2, r4, #0
    adds    r2, #14
    str     r3, [r2, #0]
A nice optimisation, but in this case it's incorrect. $r4 is aligned,
and the result of adding #14 to $r4 is an unaligned pointer.

Should gcc know better, or do I need to give it a little more
information to help it out?

gcc-help is the correct list to ask this question on. Anyway, I suspect people would be aided in helping you by seeing the source code and knowing what version of gcc you're using... I suspect you don't mark the structure as packed and as using 1 or 2 byte alignment. If you do that, then the compiler should generate the correct code, for example:

mrs $ cat t1.c
struct {
  char a[14];
  int i __attribute__((aligned(1), packed));
} s, d;

main() {
  d.i = s.i;

$ arm-gcc -O4 t1.c -S


        @ args = 0, pretend = 0, frame = 0
        @ frame_needed = 0, uses_anonymous_args = 0
        @ link register save eliminated.
        str     sl, [sp, #-4]!
        ldr     sl, .L3
        ldr     r2, .L3+4
        add     sl, pc, sl
        ldr     ip, [sl, r2]
        ldr     r0, .L3+8
        ldrh    r1, [ip, #16]
        ldr     r2, [sl, r0]
        ldrh    r3, [ip, #14]
        @ lr needed for prologue
        strh    r1, [r2, #16]   @ movhi
        strh    r3, [r2, #14]   @ movhi
        ldmfd   sp!, {sl}
        mov     pc, lr

for me. Notice the adding of 14, notice the two 16 bit moves instead of one 4 byte move. If you lie to the compiler, it will make your life rough. Telling it that it is aligned, when the data isn't aligned, is a lie.

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