stdcount if

template <class InputIterator, class UnaryPredicate>
typename iterator_traits<InputIterator>::difference_type
count_if (InputIterator first, InputIterator last, UnaryPredicate red);

效果

计算指定谓词函数为 true 的范围内的元素数

参数

first =>指向范围开头的迭代器 last =>指向范围结束的迭代器 red =>谓词函数（返回 true 或 false）

返回

指定范围内谓词函数返回 true 的元素数。

例

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

/*
    Define a few functions to use as predicates
*/

//return true if number is odd
bool isOdd(int i){
  return i%2 == 1;
}

//functor that returns true if number is greater than the value of the constructor parameter provided
class Greater {
  int _than;
public:
  Greater(int th): _than(th){}
  bool operator()(int i){
    return i > _than;
  }
};

int main(int argc, const char * argv[]) {
  
  //create a vector
  vector<int> myvec = {1,5,8,0,7,6,4,5,2,1,5,0,6,9,7};

  //using a lambda function to count even numbers
  size_t evenCount = count_if(myvec.begin(), myvec.end(), [](int i){return i % 2 == 0;}); // >= C++11
  
  //using function pointer to count odd number in the first half of the vector
  size_t oddCount = count_if(myvec.begin(), myvec.end()- myvec.size()/2, isOdd);
  
  //using a functor to count numbers greater than 5
  size_t greaterCount = count_if(myvec.begin(), myvec.end(), Greater(5));

  cout << "vector size: " << myvec.size() << endl;
  cout << "even numbers: " << evenCount << " found" << endl;
  cout << "odd numbers: " << oddCount << " found" << endl;
  cout << "numbers > 5: " << greaterCount << " found"<< endl;
  
  return 0;
}

输出

vector size: 15
even numbers: 7 found
odd numbers: 4 found
numbers > 5: 6 found