我有以下 C++17 代码,是在 x64 模式下使用 VS 2019(版本 16.8.6)编译的:
struct __declspec(align(16)) Vec2f { float v[2]; };
struct __declspec(align(16)) Vec4f { float v[4]; };
static constexpr std::uint64_t N = 100'000'000ull;
const Vec2f p{};
Vec4f acc{};
// Using virtual method:
for (std::uint64_t i = 0; i < N; ++i)
acc += foo->eval(p);
// Using function pointer:
for (std::uint64_t i = 0; i < N; ++i)
acc += eval_fn(p);
在第一个循环中,foo
is a std::shared_ptr
and eval()
是一个虚方法:
__declspec(noinline) virtual Vec4f eval(const Vec2f& p) const noexcept
{
return { p.v[0], p.v[1], p.v[0], p.v[1] };
}
在第二个循环中,eval_fn
是指向以下函数的指针:
__declspec(noinline) Vec4f eval_fn_impl(const Vec2f& p) noexcept
{
return { p.v[0], p.v[1], p.v[0], p.v[1] };
}
最后,我有两个实现operator+=
for Vec4f
:
-
一种使用显式循环实现:
Vec4f& operator+=(Vec4f& lhs, const Vec4f& rhs) noexcept
{
for (std::uint32_t i = 0; i < 4; ++i)
lhs.v[i] += rhs.v[i];
return lhs;
}
-
还有一种是用 SSE 内在函数实现的:
Vec4f& operator+=(Vec4f& lhs, const Vec4f& rhs) noexcept
{
_mm_store_ps(lhs.v, _mm_add_ps(_mm_load_ps(lhs.v), _mm_load_ps(rhs.v)));
return lhs;
}
您可以在下面找到测试的完整(独立、仅限 Windows)代码。
以下是两个循环的生成代码,以及在AMD 线程撕裂者 3970XCPU(Zen 2架构):
-
随着上证内在实施operator+=(Vec4f&, const Vec4f&)
:
// Using virtual method: 649 ms
$LL4@main:
mov rax, QWORD PTR [rdi] // fetch vtable base pointer (rdi = foo)
lea r8, QWORD PTR p$[rsp] // r8 = &p
lea rdx, QWORD PTR $T3[rsp] // not sure what $T3 is (some kind of temporary, but why?)
mov rcx, rdi // rcx = this
call QWORD PTR [rax] // foo->eval(p)
addps xmm6, XMMWORD PTR [rax]
sub rbp, 1
jne SHORT $LL4@main
// Using function pointer: 602 ms
$LL7@main:
lea rdx, QWORD PTR p$[rsp] // rdx = &p
lea rcx, QWORD PTR $T2[rsp] // same question as above
call rbx // eval_fn(p)
addps xmm6, XMMWORD PTR [rax]
sub rsi, 1
jne SHORT $LL7@main
-
随着显式循环实施operator+=(Vec4f&, const Vec4f&)
:
// Using virtual method: 167 ms [3.5x to 4x FASTER!]
$LL4@main:
mov rax, QWORD PTR [rdi]
lea r8, QWORD PTR p$[rsp]
lea rdx, QWORD PTR $T5[rsp]
mov rcx, rdi
call QWORD PTR [rax]
addss xmm9, DWORD PTR [rax]
addss xmm8, DWORD PTR [rax+4]
addss xmm7, DWORD PTR [rax+8]
addss xmm6, DWORD PTR [rax+12]
sub rbp, 1
jne SHORT $LL4@main
// Using function pointer: 600 ms
$LL7@main:
lea rdx, QWORD PTR p$[rsp]
lea rcx, QWORD PTR $T4[rsp]
call rbx
addps xmm6, XMMWORD PTR [rax]
sub rsi, 1
jne SHORT $LL7@main
(据我所知,在 AMD Zen 2 架构上,addss
and addps
指令有 3 个周期的延迟,最多可以同时执行两条这样的指令。)
让我困惑的情况是使用虚拟方法和显式循环实现时operator+=
:
为什么它比其他三个变体快 3.5 倍到 4 倍?
这里有哪些相关的建筑效果?循环后续迭代中寄存器之间的依赖性更少?或者缓存方面运气不好?
完整源代码:
#include <Windows.h>
#include <cstdint>
#include <cstdio>
#include <memory>
#include <xmmintrin.h>
struct __declspec(align(16)) Vec2f
{
float v[2];
};
struct __declspec(align(16)) Vec4f
{
float v[4];
};
Vec4f& operator+=(Vec4f& lhs, const Vec4f& rhs) noexcept
{
#if 0
_mm_store_ps(lhs.v, _mm_add_ps(_mm_load_ps(lhs.v), _mm_load_ps(rhs.v)));
#else
for (std::uint32_t i = 0; i < 4; ++i)
lhs.v[i] += rhs.v[i];
#endif
return lhs;
}
std::uint64_t get_timer_freq()
{
LARGE_INTEGER frequency;
QueryPerformanceFrequency(&frequency);
return static_cast<std::uint64_t>(frequency.QuadPart);
}
std::uint64_t read_timer()
{
LARGE_INTEGER count;
QueryPerformanceCounter(&count);
return static_cast<std::uint64_t>(count.QuadPart);
}
struct Foo
{
__declspec(noinline) virtual Vec4f eval(const Vec2f& p) const noexcept
{
return { p.v[0], p.v[1], p.v[0], p.v[1] };
}
};
using SampleFn = Vec4f (*)(const Vec2f&);
__declspec(noinline) Vec4f eval_fn_impl(const Vec2f& p) noexcept
{
return { p.v[0], p.v[1], p.v[0], p.v[1] };
}
__declspec(noinline) SampleFn make_eval_fn()
{
return &eval_fn_impl;
}
int main()
{
static constexpr std::uint64_t N = 100'000'000ull;
const auto timer_freq = get_timer_freq();
const Vec2f p{};
Vec4f acc{};
{
const auto foo = std::make_shared<Foo>();
const auto start_time = read_timer();
for (std::uint64_t i = 0; i < N; ++i)
acc += foo->eval(p);
std::printf("foo->eval: %llu ms\n", 1000 * (read_timer() - start_time) / timer_freq);
}
{
const auto eval_fn = make_eval_fn();
const auto start_time = read_timer();
for (std::uint64_t i = 0; i < N; ++i)
acc += eval_fn(p);
std::printf("eval_fn: %llu ms\n", 1000 * (read_timer() - start_time) / timer_freq);
}
return acc.v[0] + acc.v[1] + acc.v[2] + acc.v[3] > 0.0f ? 1 : 0;
}