Question

    int nums[] = {7, 6, 12, 9, 29, 1, 67, 3, 3, 8, 9, 77};
    std::vector<int> vecInts(&nums[0], &nums[0] + sizeof(nums)/sizeof(nums[0]));

    int countBoost = 0;

    // (i > 5 && i <=10)
    countBoost = std::count_if(vecInts.begin(), vecInts.end(),
                      boost::bind(std::logical_and<bool>(), 
                                  boost::bind(std::greater<int>(),    _1, 5),
                                         boost::bind(std::less_equal<int>(), _1, 10))
                          );

现在，我需要使用纯STL实现相同的逻辑。我怎么能这样做？

我尝试过以下代码，但不起作用：

int countSTL   = std::count_if(vecInts.begin(), vecInts.end(),
                           std::logical_and<bool>(std::bind2nd(std::greater<int>(), 5), std::bind2nd(std::less_equal<int>(), 10))                               
                          );

谢谢

//更新//

In Effective STL Item 43, Meyers indicates as follows:

vector<int>::iterator i = find_if(v.begin(), v.end(),
           compose2(logical_and<bool>(), bind2nd(greater<int>(), x),
                                         bind2nd(less<int>(), y)));

但是compose2不是标准的函数对象适配器。

Answer 1

使用“纯”C ++ 03标准 - 你只能使用额外的bool数组：将bind2nd(greater<int>(), x)的所有结果存储到一个bool数组，第二个数组中的less相同。 logical_and结果为第三个数组。对于动态大小 - 使用std :: vector而不是简单的原始数组。或者只是从http://www.sgi.com/tech/stl/stl_function.h复制（窃取）SGI STL compose2<>的实现。

int main() {
    int nums[] = {7, 6, 12, 9, 29, 1, 67, 3, 3, 8, 9, 77};
    const size_t NUMS_SIZE = sizeof(nums) / sizeof(*nums);
    bool nums_greater[NUMS_SIZE];
    bool nums_less[NUMS_SIZE];
    bool nums_greater_and_less[NUMS_SIZE];

    int x = 3;
    int y = 20;
    transform(nums, nums + NUMS_SIZE, nums_greater, bind2nd(greater<int>(), x));
    transform(nums, nums + NUMS_SIZE, nums_less, bind2nd(less<int>(), y));
    transform (nums_greater, nums_greater+NUMS_SIZE, nums_less, nums_greater_and_less,
               logical_and<bool>() );

    int countBoost = 0;

    countBoost = count(nums_greater_and_less, nums_greater_and_less + NUMS_SIZE, true);

    cout << countBoost  << endl;
}

使用std :: logical_and来组合两个条件

1 个答案: