在各种情况下，我观​​察到C
++中的链表迭代始终比Go中的慢10-15％。我在解决堆栈溢出这个神秘的第一次尝试是在这里。我编写的示例有问题，因为：

1）由于堆分配，内存访问是不可预测的，并且

2）因为没有实际的工作要做，所以一些人的编译器正在优化主循环。

为了解决这些问题，我有一个新程序，其中包含C 和Go的实现。C
版本花费1.75秒，而Go版本花费1.48秒。这次，我在计时开始之前进行了一次大堆分配，并使用它来操作对象池，并从中释放并获取链表的节点。这样，两种实现之间的内存访问应该完全相似。

希望这可以使奥秘重现！

C ++：

#include <iostream>
#include <sstream>
#include <fstream>
#include <string>
#include <vector>
#include <boost/timer.hpp>

using namespace std;

struct Node {
    Node *next; // 8 bytes
    int age;   // 4 bytes
};

// Object pool, where every free slot points to the previous free slot
template<typename T, int n>
struct ObjPool
{
    typedef T*       pointer;
    typedef pointer* metapointer;

    ObjPool() :
        _top(NULL),
        _size(0)
    {
        pointer chunks = new T[n];
        for (int i=0; i < n; i++) {
            release(&chunks[i]);
        }
    }

    // Giver an available pointer to the object pool
    void release(pointer ptr)
    {
        // Store the current pointer at the given address
        *(reinterpret_cast<metapointer>(ptr)) = _top;

        // Advance the pointer
        _top = ptr;

        // Increment the size
        ++_size;
    }

    // Pop an available pointer off the object pool for program use
    pointer acquire(void)
    {
        if(_size == 0){throw std::out_of_range("");}

        // Pop the top of the stack
        pointer retval = _top;

        // Step back to the previous address
        _top = *(reinterpret_cast<metapointer>(_top));

        // Decrement the size
        --_size;

        // Return the next free address
        return retval;
    }

    unsigned int size(void) const {return _size;}

protected:
    pointer _top;

    // Number of free slots available
    unsigned int _size;
};

Node *nodes = nullptr;
ObjPool<Node, 1000> p;

void processAge(int age) {
    // If the object pool is full, pop off the head of the linked list and release
    // it from the pool
    if (p.size() == 0) {
        Node *head = nodes;
        nodes = nodes->next;
        p.release(head);
    }

    // Insert the new Node with given age in global linked list. The linked list is sorted by age, so this requires iterating through the nodes.
    Node *node = nodes;
    Node *prev = nullptr;
    while (true) {
        if (node == nullptr || age < node->age) {
            Node *newNode = p.acquire();
            newNode->age = age;
            newNode->next = node;

            if (prev == nullptr) {
                nodes = newNode;
            } else {
                prev->next = newNode;
            }

            return;
        }

        prev = node;
        node = node->next;
    }
}

int main() {
    Node x = {};
    std::cout << "Size of struct: " << sizeof(x) << "\n"; // 16 bytes

    boost::timer t;
    for (int i=0; i<1000000; i++) {
        processAge(i);
    }

    std::cout << t.elapsed() << "\n";
}

走：

package main

import (
    "time"
    "fmt"
    "unsafe"
)

type Node struct {
    next *Node // 8 bytes
    age int32 // 4 bytes
}

// Every free slot points to the previous free slot
type NodePool struct {
    top *Node
    size int
}

func NewPool(n int) NodePool {
    p := NodePool{nil, 0}
    slots := make([]Node, n, n)
    for i := 0; i < n; i++ {
        p.Release(&slots[i])
    }

    return p
}

func (p *NodePool) Release(l *Node) {
    // Store the current top at the given address
    *((**Node)(unsafe.Pointer(l))) = p.top
    p.top = l
    p.size++
}

func (p *NodePool) Acquire() *Node {
    if p.size == 0 {
        fmt.Printf("Attempting to pop from empty pool!\n")
    }
    retval := p.top

    // Step back to the previous address in stack of addresses
    p.top = *((**Node)(unsafe.Pointer(p.top)))
    p.size--
    return retval
}

func processAge(age int32) {
    // If the object pool is full, pop off the head of the linked list and release
    // it from the pool
    if p.size == 0 {
        head := nodes
        nodes = nodes.next
        p.Release(head)
    }

    // Insert the new Node with given age in global linked list. The linked list is sorted by age, so this requires iterating through the nodes.
    node := nodes
    var prev *Node = nil
    for true {
        if node == nil || age < node.age {
            newNode := p.Acquire()
            newNode.age = age
            newNode.next = node

            if prev == nil {
                nodes = newNode
            } else {
                prev.next = newNode
            }
            return
        }

        prev = node
        node = node.next
    }
}

// Linked list of nodes, in ascending order by age
var nodes *Node = nil
var p NodePool = NewPool(1000)

func main() {
    x := Node{};
    fmt.Printf("Size of struct: %d\n", unsafe.Sizeof(x)) // 16 bytes

    start := time.Now()
    for i := 0; i < 1000000; i++ {
        processAge(int32(i))
    }

    fmt.Printf("Time elapsed: %s\n", time.Since(start))
}

输出：

clang++ -std=c++11 -stdlib=libc++ minimalPool.cpp -O3; ./a.out
Size of struct: 16
1.7548

go run minimalPool.go
Size of struct: 16
Time elapsed: 1.487930629s