stdfind

template <class InputIterator, class T>
InputIterator find (InputIterator first, InputIterator last, const T& val);

效果

查詢範圍內第一次出現的 val [first，last）

引數

first =>指向範圍開頭的迭代器 last =>指向範圍結束的迭代器 val =>要在範圍內查詢的值

返回

迭代器指向範圍內等於（==）到 val 的第一個元素，如果未找到 val，則迭代器指向 last。

例

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

using namespace std;

int main(int argc, const char * argv[]) {

  //create a vector
  vector<int> intVec {4, 6, 8, 9, 10, 30, 55,100, 45, 2, 4, 7, 9, 43, 48};

  //define iterators
  vector<int>::iterator  itr_9; 
  vector<int>::iterator  itr_43; 
  vector<int>::iterator  itr_50; 

  //calling find
  itr_9 = find(intVec.begin(), intVec.end(), 9); //occurs twice
  itr_43 = find(intVec.begin(), intVec.end(), 43); //occurs once

  //a value not in the vector
  itr_50 = find(intVec.begin(), intVec.end(), 50); //does not occur

  cout << "first occurence of: " << *itr_9 << endl;
  cout << "only occurence of: " << *itr_43 << Lendl;

  /*
    let's prove that itr_9 is pointing to the first occurence
    of 9 by looking at the element after 9, which should be 10 
    not 43
  */
  cout << "element after first 9: " << *(itr_9 + 1) << ends;

  /*
    to avoid dereferencing intVec.end(), lets look at the 
    element right before the end
  */
  cout << "last element: " << *(itr_50 - 1) << endl;

  return 0;
}

輸出

first occurence of: 9
only occurence of: 43
element after first 9: 10
last element: 48