现有阵列周围的STL非复制包装器?

您可以直接在常规的C样式数组上调用许多STL算法——它们就是为此而设计的。例如。,:

int ary[100];
// init ...

std::sort(ary, ary+100); // sorts the array
std::find(ary, ary+100, pred); find some element

可以使用内联函数模板,这样就不必复制数组索引

template <typename T, int I>
inline T * array_begin (T (&t)[I])
{
  return t;
}

template <typename T, int I>
inline T * array_end (T (&t)[I])
{
  return t + I;
}

void foo ()
{
  int array[100];
  std::find (array_begin (array)
      , array_end (array)
      , 10);
}

所有STL算法都使用迭代器。
指针是对象数组的有效迭代器。

请注意

#include <algorithm>
#include <iostream>
#include <iterator>

int main()
{
    int   data[] = {4,3,7,5,8};
    std::sort(data,data+5);

    std::copy(data,data+5,std::ostream_iterator<int>(std::cout,"\t"));
}

你可以用 Boost.Array

使用数组:

int a[100];
for (int i = 0; i < 100; ++i)
    a[i] = 0;

使用boost.Array:

boost::array<int,100> a;
for (boost::array<int,100>::iterator i = a.begin(); i != a.end(); ++i)
    *i = 0;

更新: 使用C++11,您现在可以使用 std::array

指针是迭代器的有效模型:

struct Bob
{ int val; };

bool operator<(const Bob& lhs, const Bob& rhs)
{ return lhs.val < rhs.val; }

// let's do a reverse sort
bool pred(const Bob& lhs, const Bob& rhs)
{ return lhs.val > rhs.val; }

bool isBobNumberTwo(const Bob& bob) { return bob.val == 2; }

int main()
{
    Bob bobs[4]; // ok, so we have 4 bobs!
    const size_t size = sizeof(bobs)/sizeof(Bob);
    bobs[0].val = 1; bobs[1].val = 4; bobs[2].val = 2; bobs[3].val = 3;

    // sort using std::less<Bob> wich uses operator <
    std::sort(bobs, bobs + size);
    std::cout << bobs[0].val << std::endl;
    std::cout << bobs[1].val << std::endl;
    std::cout << bobs[2].val << std::endl;
    std::cout << bobs[3].val << std::endl;

    // sort using pred
    std::sort(bobs, bobs + size, pred);
    std::cout << bobs[0].val << std::endl;
    std::cout << bobs[1].val << std::endl;
    std::cout << bobs[2].val << std::endl;
    std::cout << bobs[3].val << std::endl;

    //Let's find Bob number 2
    Bob* bob = std::find_if(bobs, bobs + size, isBobNumberTwo);
    if (bob->val == 2)
        std::cout << "Ok, found the right one!\n";
    else 
        std::cout << "Whoops!\n";

    return 0;
}