c++ - std::vector 是否在内存中有连续的数据？

Question

class Wrapper {
public:
    // some functions operating on the value_
    __m128i value_;
};

int main() {
    std::vector<Wrapper> a;
    a.resize(100);
}

vector a 中的 Wrapper 对象的 value_ 属性是否始终占据连续的内存且 __m128i 值之间没有任何间隙 ?

我的意思是:

[128 bit for 1st Wrapper][no gap here][128bit for 2nd Wrapper] ...

到目前为止，这似乎适用于 g++ 和我正在使用的 Intel cpu，以及 gcc godbolt。

由于 Wrapper 对象中只有一个 __m128i 属性，这是否意味着编译器始终 不需要在内存中添加任何类型的填充？ ( Memory layout of vector of POD objects )

测试代码1:

#include <iostream>
#include <vector>
#include <x86intrin.h>

int main()
{
  static constexpr size_t N = 1000;
  std::vector<__m128i> a;
  a.resize(1000);
  //__m128i a[1000];
  uint32_t* ptr_a = reinterpret_cast<uint32_t*>(a.data());
  for (size_t i = 0; i < 4*N; ++i)
    ptr_a[i] = i;
  for (size_t i = 1; i < N; ++i){
    a[i-1] = _mm_and_si128 (a[i], a[i-1]);
  }
  for (size_t i = 0; i < 4*N; ++i)
    std::cout << ptr_a[i];
}

警告:

warning: ignoring attributes on template argument 
'__m128i {aka __vector(2) long long int}'
[-Wignored-attributes]

程序集(gcc god bolt):

.L9:
        add     rax, 16
        movdqa  xmm1, XMMWORD PTR [rax]
        pand    xmm0, xmm1
        movaps  XMMWORD PTR [rax-16], xmm0
        cmp     rax, rdx
        movdqa  xmm0, xmm1
        jne     .L9

我猜这意味着数据是连续的，因为循环只是将 16 个字节添加到它在循环的每个循环中读取的内存地址。它使用 pand 进行按位与。

测试代码2:

#include <iostream>
#include <vector>
#include <x86intrin.h>
class Wrapper {
public:
    __m128i value_;
    inline Wrapper& operator &= (const Wrapper& rhs)
    {
        value_ = _mm_and_si128(value_, rhs.value_);
    }
}; // Wrapper
int main()
{
  static constexpr size_t N = 1000;
  std::vector<Wrapper> a;
  a.resize(N);
  //__m128i a[1000];
  uint32_t* ptr_a = reinterpret_cast<uint32_t*>(a.data());
  for (size_t i = 0; i < 4*N; ++i) ptr_a[i] = i;
  for (size_t i = 1; i < N; ++i){
    a[i-1] &=a[i];
    //std::cout << ptr_a[i];
  }
  for (size_t i = 0; i < 4*N; ++i)
    std::cout << ptr_a[i];
}

程序集(gcc god bolt)

.L9:
        add     rdx, 2
        add     rax, 32
        movdqa  xmm1, XMMWORD PTR [rax-16]
        pand    xmm0, xmm1
        movaps  XMMWORD PTR [rax-32], xmm0
        movdqa  xmm0, XMMWORD PTR [rax]
        pand    xmm1, xmm0
        movaps  XMMWORD PTR [rax-16], xmm1
        cmp     rdx, 999
        jne     .L9

看起来也没有填充。 rax 每一步增加 32，即 2 x 16。那个额外的 add rdx,2 肯定不如测试代码 1 的循环。

测试自动矢量化

#include <iostream>
#include <vector>
#include <x86intrin.h>

int main()
{
  static constexpr size_t N = 1000;
  std::vector<__m128i> a;
  a.resize(1000);
  //__m128i a[1000];
  uint32_t* ptr_a = reinterpret_cast<uint32_t*>(a.data());
  for (size_t i = 0; i < 4*N; ++i)
    ptr_a[i] = i;
  for (size_t i = 1; i < N; ++i){
    a[i-1] = _mm_and_si128 (a[i], a[i-1]);
  }
  for (size_t i = 0; i < 4*N; ++i)
    std::cout << ptr_a[i];
}

程序集(god bolt):

.L21:
        movdqu  xmm0, XMMWORD PTR [r10+rax]
        add     rdi, 1
        pand    xmm0, XMMWORD PTR [r8+rax]
        movaps  XMMWORD PTR [r8+rax], xmm0
        add     rax, 16
        cmp     rsi, rdi
        ja      .L21

...我只是不知道对于英特尔 cpu 和 g++/英特尔 c++ 编译器是否总是如此/(在此处插入编译器名称)...

Answer 1

不能保证 class Wrapper 的末尾不会有填充，只有它的开头不会有填充。

根据C++11标准:

9.2 Class members [ class.mem ]

20 A pointer to a standard-layout struct object, suitably converted using a reinterpret_cast, points to its initial member (or if that member is a bit-field, then to the unit in which it resides) and vice versa. [ Note: There might therefore be unnamed padding within a standard-layout struct object, but not at its beginning, as necessary to achieve appropriate alignment. — end note ]

同样在sizeof下:

5.3.3 Sizeof [ expr.sizeof ]

2 When applied to a reference or a reference type, the result is the size of the referenced type. When applied to a class, the result is the number of bytes in an object of that class including any padding required for placing objects of that type in an array.