我在这里做错了什么？我得到 4 个零而不是:

2
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6
8

#include <iostream>
#include <chrono>

extern "C" double *addVec(double *C, double *A, double *B, size_t &N);

int main()
{
    size_t N = 1 << 2;
    size_t reductions = N / 4;

    double *A = (double*)_aligned_malloc(N*sizeof(double), 32);
    double *B = (double*)_aligned_malloc(N*sizeof(double), 32);
    double *C = (double*)_aligned_malloc(N*sizeof(double), 32);

    for (size_t i = 0; i < N; i++)
    {
        A[i] = double(i + 1);
        B[i] = double(i + 1);
    }

    auto start = std::chrono::high_resolution_clock::now();

        double *out = addVec(C, A, B, reductions);

    auto finish = std::chrono::high_resolution_clock::now();

    for (size_t i = 0; i < N; i++)
    {
        std::cout << out[i] << std::endl;
    }

    std::cout << "\n\n";

    std::cout << std::chrono::duration_cast<std::chrono::nanoseconds>(finish - start).count() << " ns\n";

    std::cin.get();

    _aligned_free(A);
    _aligned_free(B);
    _aligned_free(C);

    return 0;
}

.data
; C -> RCX
; A -> RDX
; B -> r8
; N -> r9
.code
    addVec proc
        ;xor rbx, rbx
        align 16
        ;aIn:
            vmovapd ymm0, ymmword ptr [rdx]
            ;vmovapd ymm1, ymmword ptr [rdx + rbx + 4]
            vmovapd ymm2, ymmword ptr [r8]
            ;vmovapd ymm3, ymmword ptr [r8 + rbx + 4]

            vaddpd ymm0, ymm2, ymm3

            vmovapd ymmword ptr [rcx], ymm3
        ;inc rbx
        ;cmp rbx, qword ptr [r9]
        ;jl aIn
        mov rax, rcx    ; return the address of the output vector
    ret
    addVec endp
end

.data
;// C -> RCX
;// A -> RDX
;// B -> r8
.code
    addVec proc
        ; One cycle 8 micro-op
            vmovapd ymm0, ymmword ptr [rdx]     ; 1 port
            vmovapd ymm1, ymmword ptr [rdx + 32]; 1 port
            vmovapd ymm2, ymmword ptr [r8]      ; 1 port
            vmovapd ymm3, ymmword ptr [r8 + 32] ; 1 port
            vfmadd231pd ymm0, ymm2, ymm4        ; 1 port
            vfmadd231pd ymm1, ymm3, ymm4        ; 1 port
            vmovapd ymmword ptr [rcx], ymm0     ; 1 port
            vmovapd ymmword ptr [rcx + 32], ymm1; 1 port

        ; Return the address of the output vector
        mov rax, rcx                            ; 1 port ?
    ret
    addVec endp
end

.data
;// C -> RCX
;// A -> RDX
;// B -> r8
.code
    addVec proc
        ;align 16
        ; One cycle 5 micro-op ?
        vmovapd ymm0, ymmword ptr [rdx]     ; 1 port
        vmovapd ymm1, ymmword ptr [r8]      ; 1 port
        vfmadd231pd ymm0, ymm1, ymm2        ; 1 port
        vmovapd ymmword ptr [rcx], ymm0     ; 1 port

        ; Return the address of the output vector
        mov rax, rcx                        ; 1 port ?
    ret
    addVec endp
end

最佳答案

您的代码得到错误结果的原因是您的程序集中的语法向后。

您正在使用 Intel 语法，其中目标应位于源之前。所以在你原来的 .asm 代码中你应该改变

vaddpd ymm0, ymm2, ymm3

 vaddpd ymm3, ymm2, ymm0

extern "C" double *addVec(double * __restrict C, double * __restrict A, double * __restrict B, size_t &N) {
    __m256d x = _mm256_load_pd((const double*)A);
    __m256d y = _mm256_load_pd((const double*)B);
    __m256d z = _mm256_add_pd(x,y);
    _mm256_store_pd((double*)C, z);
    return C;
}

vmovapd ymm0, YMMWORD PTR [rdx]
mov     rax, rcx
vaddpd  ymm0, ymm0, YMMWORD PTR [r8]
vmovapd YMMWORD PTR [rcx], ymm0
vzeroupper
ret

256-load + 256-FMA    //one fused µop using two ports
256-load + 256-FMA    //one fused µop using two ports
256-store             //one µop using two ports
64-bit add + jump     //one µop using one port

关于c++ - 使用 AVX2 在程序集 x86_64 中添加两个 vector 加上技术说明，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/26820662/