Why is the generated assembly reordered when using the built-in functions?

Question

I played a little with the built-in functions, since I needed a complexity function O (1)similar memcmp()to that for a fixed input size. I ended up writing this:

#include <stdint.h>
#include <emmintrin.h>

int64_t f (int64_t a[4], int64_t b[4]) {
    __m128i *x = (void *) a, *y = (void *) b, r[2], t;
    int64_t *ret = (void *) &t;

    r[0] = _mm_xor_si128(x[0], y[0]);
    r[1] = _mm_xor_si128(x[1], y[1]);
    t = _mm_or_si128(r[0], r[1]);


    return (ret[0] | ret[1]);
}

which when compiled turns into this:

f:
    movdqa  xmm0, XMMWORD PTR [rdi]
    movdqa  xmm1, XMMWORD PTR [rdi+16]
    pxor    xmm0, XMMWORD PTR [rsi]
    pxor    xmm1, XMMWORD PTR [rsi+16]
    por xmm0, xmm1
    movq    rdx, xmm0
    pextrq  rax, xmm0, 1
    or  rax, rdx
    ret

http://goo.gl/EtovJa (Godbolt Compiler Explorer)

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After that, I became curious to find out if I really needed to use internal functions or if I only needed types, and I could just use regular operators. Then I modified the code above (only three SSE lines) and as a result:

#include <stdint.h>
#include <emmintrin.h>

int64_t f (int64_t a[4], int64_t b[4]) {
    __m128i *x = (void *) a, *y = (void *) b, r[2], t;
    int64_t *ret = (void *) &t;

    r[0] = x[0] ^ y[0];
    r[1] = x[1] ^ y[1];
    t = r[0] | r[1];


    return (ret[0] | ret[1]);
}

which compiles instead:

f:
    movdqa  xmm0, XMMWORD PTR [rdi+16]
    movdqa  xmm1, XMMWORD PTR [rdi]
    pxor    xmm0, XMMWORD PTR [rsi+16]
    pxor    xmm1, XMMWORD PTR [rsi]
    por xmm0, xmm1
    movq    rdx, xmm0
    pextrq  rax, xmm0, 1
    or  rax, rdx
    ret

http://goo.gl/oDHF3z (Godbolt Compiler Explorer)

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. GCC .

Answer 1

TL; DR. , .

() clang, intrinsics , IR ( , LLVM), IR , IR .

clang gcc, . - , .

, , . gcc clang. Clang movd , llvm IR .

GCC, , , . - , intrinsics (not function).