C++内存对齐是正确还是低效？

Question

我测试了这段代码,试图找出为新运算符实际保留了多少内存c ++.

#include<iostream>
using namespace std;
int main() {

  cout << "alignment of " << alignof(int) << endl;
  int *intP1 = new int;
  *intP1 = 100;
  cout << "address of intP1 " << intP1 << endl;
  int *intP2 = new int;
  *intP2 = 999;
  cout << "address of intP2 " << intP2 << endl;
  int *intP3 = new int;
  cout << "address of intP3 " << intP3 << endl;
  *intP3 = 333;

  cout << endl;
  cout << (reinterpret_cast<char *>(intP3)-reinterpret_cast<char *>(intP2)) << endl;
  cout << intP3-intP2 << endl;
  cout << endl;

  cout << *(intP1) << endl;
  cout << *(intP1+4) << endl;
  cout << *(intP1+8) << endl;
  cout << *(intP1+16) << endl;
  delete intP1;
  delete intP2;
  delete intP3;
  return 0;
}

在使用-std = c ++ 11标志编译代码并运行它之后,这是我从x86_64机器获得的.

    alignment of int4
    address of intP1 = 0xa59010
    address of intP2 = 0xa59030
    address of intP3 = 0xa59050

    the distance of intP3 and intP2 = 32

    intP1 value = 100
    is this a padding value = 0
    intP2 value = 999
    intP3 value = 333

似乎当使用new为一个整数分配一个4字节的内存时,它实际上保留了32个字节的块,这是8个整数的总空间.根据c ++对齐的解释,对于64位机器,内存在16个字节上对齐,为什么这里的距离是32个字节？

有人可以帮我解决这个问题吗？提前致谢.

Answer 1

它与对齐无关 - 它是内部内存分配器工作方式的额外开销.通常,每个内存块在其前面和/或后面都有额外的隐藏信息,用于维护堆的结构.究竟有多少开销会因平台和实现而异.

例如,Doug Lea的malloc额外开销为每个分配4-8个字节(32位指针)或8-16个字节(64位指针),最小分配大小为16个字节(32位)或32个字节( 64位).这意味着即使是1字节的分配,内存分配器也需要总共16个字节的跟踪开销.