C++ - 迭代从 find_if 返回的 std::vector<>

Question

I'm learning C++, so I feel like this should be a very simple answer - but I can't seem to find it. So I apologize in advance if it's naive.

I have a std::vector<int> of of values, and I am trying to find the indices of the odd values.

I am following the code from here :

(repeated below):

// find_if example
#include <iostream>     // std::cout
#include <algorithm>    // std::find_if
#include <vector>       // std::vector

bool IsOdd (int i) {
  return ((i%2)==1);
}

int main () {
  std::vector<int> myvector;

  myvector.push_back(10);
  myvector.push_back(25);
  myvector.push_back(40);
  myvector.push_back(55);

  std::vector<int>::iterator it = std::find_if (myvector.begin(), myvector.end(), IsOdd);
  std::cout << "The first odd value is " << *it << '\n';

  return 0;
}

This example prints the first odd value. How could I extend this to give me the index values for each of the odd values in myvector ? Is this the correct approach?

Answer 1

// find_if example
#include <iostream>     // std::cout
#include <algorithm>    // std::find_if
#include <vector>       // std::vector

bool IsOdd (int i) {
  return ((i%2)==1);
}

int main () {
  std::vector<int> myvector;

  myvector.push_back(10);
  myvector.push_back(25);
  myvector.push_back(40);
  myvector.push_back(55);


  std::vector<int>::iterator it = std::find_if (myvector.begin(), myvector.end(), IsOdd); 
  std::cout << "ODD values are: " << std::endl;     

  while(it != myvector.end() ){

    std::cout << *it  << " in position " << (it - myvector.begin())  << '\n';
    it = std::find_if (++it, myvector.end(), IsOdd); 
  }
  return 0;
}

EDIT: Changed it+1 to ++it see @David Rodríguez - dribeas comment below.

Answer 2

You can increment it and use it as a starting point to iterate further:

std::cout << "odd values: ";
auto it = myvector.begin();
while(it != myvector.end())
{
   it = std::find_if (it, myvector.end(), IsOdd);
   if(it == myvector.end()) break;
   std::cout << *it << ' ';
   ++it;
}
std::cout << endl;

A much more algorithm oriented approach, makes use of copy_if , having an output vector as a result container:

std::vector<int> results;
std::copy_if(myvector.begin(), myvector.end(), std::back_inserter(results), IsOdd);

Now results contains the odd values. (Note the back:inserter is in the <iterator> header)

Answer 3

You can find the index of a vector iterator (and, more generally, any random-access iterator) by subtracting the start of the sequence:

std::cout << "The index is " << (it - myvector.begin()) << '\n';

Even more generally, there is a std::distance function which can give you the distance between forward iterators. You could use that, for example, if your container were a list ; but you probably wouldn't want to, since it would be much slower.

To find all the odd numbers, you'll need a loop to call find again, starting from the element after the one you just found.

Answer 4

You'll need a loop . The iterator-algorithm design of the standard library makes this pretty easy:

#include <iterator>

for (auto it = myvector.begin();
     (it = std::find_if(it, myvector.end(), IsOdd)) != myvector.end(); )
{
    std::cout << *it << " at index " << std::distance(myvector.begin(), it) << "\n";
}

Answer 5

Change these two lines:

std::vector<int>::iterator it = std::find_if (myvector.begin(), myvector.end(), IsOdd);
std::cout << "The first odd value is " << *it << '\n';

into something like:

std::vector<int>::iterator it = std::find_if (myvector.begin(), myvector.end(), IsOdd);
while ( it != myvector.end() ) {
    std::cout << "The next odd value is " << *it << '\n';
    it = std::find_if (++it, myvector.end(), IsOdd);
}

Answer 6

A nice compact solution could be:

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

int main() {
  std::vector<int> const v{1,4,9,11,2,7,8};
  std::cout << "Odd values at indices";
  for(auto b=begin(v), i=b, e=end(v);
      (i=find_if(i,e,[](int a){return a%2==1;})) != e;
      ++i)
    std::cout << ' ' << distance(b,i);
  std::cout.flush();
}