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Sandor Dargo
Sandor Dargo

Posted on • Updated on • Originally published at sandordargo.com

Replace std::find_if in 80% of the cases

If you've been following the posts on this blog, you know that I'm a big fan of using standard algorithms in any programming language, in particular in C++.

They contain fewer bugs, in general, they have better performance and the standard algorithms are more expressive.

The last point on expressiveness is very important to me and after I saw a not-so-ideal example of using std::find_if in our codebase, I wanted to have a deeper look.

So I went through all our usages of find_if and I found that it was only used in a proper way in about 20% of all the cases.

This means that the Pareto principle applies here too. In 80% of the cases, std::find_if should not have been used.

But what else should have been used? And why?

I brought some simplified examples.

Is there any such element?

Here is the first example:

std::vector numbers {1, 3, 5, 7, 9};

return numbers.end()
           != std::find_if(numbers.begin(), numbers.end(), [](int number) { return number % 2 == 1; });
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You might also see a close variant of the above example in your code base. Sometimes, there is a temporary variable to store the returned value of find_if, even if it's used only once:

auto foundElement = std::find_if(numbers.begin(), numbers.end(), [](int number) { return number % 2 == 1; });

return numbers.end() != foundElement;
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So what goes on here?

First of all, what does find_if return?

It returns an iterator to the first element of the searched range that satisfies the condition. If there is no such item, it returns an iterator pointing beyond the last element, in other words, to end().

The function's return value in the above examples is a boolean, we simply compare whether find_if returns anything else than the end() of the examined collection. In other words, it checks whether the find _if returns an iterator to any of the elements in numbers. Yet in other words, we check if any of numbers's elements satisfy the condition passed to find_if.

Alright, this last sentence should give us the hint. We can replace the above expression with std::any_of:

return std::any_of(numbers.begin(), numbers.end(), [](int number) { return number % 2 == 1; });
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What did we gain? We have a comparison less and potentially a temporary variable less as well. At the same time, our code is shorter, more expressive and we didn't even have to touch the lambda we wrote.

There is no such element!

A bit different, yet similar example:

auto aPotentialItem =
  std::find_if(items->begin(), item->end(), [&iName](const Item& anItem) {
    return inItem._name == iName;
  });
return (aPotentialItem == items->end()) ? nullptr : &(*aPotentialItem);
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In this example, we don't use != as a comparison between the end() of the collection and the return value of find_if, but == instead. This means that we check whether there is no element in a given range complying with our condition.

In other words, we check whether none of the elements satisfy our condition.

Yet, we cannot replace find_if in this example with none_of, given that we'd have to look up aPotentialItem anyway for the other case. (Thanks a lot for your comment cbuchart!)

At the same time, find_if sometimes can be replaced with none_of, when you're only looking for the result of the comparison:

std::vector numbers {1, 3, 5, 7, 9};

return std::find_if(numbers.begin(), numbers.end(), [](int number) {
    return number % 2 == 1;
  }) == numbers.end();
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In the above example, we can simplify std::find_if with std::none_of.


std::vector numbers {1, 3, 5, 7, 9, 8};

return std::none_of(numbers.begin(), numbers.end(), [](int number) {
    return number % 2 == 0;
  });
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All of the items are the same?

A slightly different case is when you use find_if_not and you compare if the returned iterator is the end() of the container.

std::vector numbers {1, 3, 5, 7, 9};

if (std::find_if_not(numbers.begin(), numbers.end(), [](int i) { return i % 2 == 0;}) == numbers.end()) {
  // do something
}
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In this case, you're looking for if there is no element matching the predicate.

We can replace it with all_of and the result will be much more readable:

std::vector numbers {1, 3, 5, 7, 9};

if (std::all_of(numbers.begin(), numbers.end(), [](int i) { return i % 2 == 0;})) {
  // do something
}
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So what to do?

Based on the cases I saw, I came up with this rule of thumb for the cases when we don't want to dereference the returned iterator, but we only use it for comparison :

  • if the result of find_if is compared using != end(), use any_of
  • if the result of find_if is compared using == end(), use none_of
  • if the results of find_if_not is compared using == end() use all_of

Keep find_if only if you want to interact with the object pointed by the returned iterator. If the outcome is just a boolean, like in the above examples, you have an alternative still in the standard library.

Conclusion

This was a short post on how to use the C++ standard algorithms in a better way than often it is used. std::find_if is often misused, probably because it's something more people know about than the alternatives.

In the vast majority of the cases I saw, it can be replaced either with std::any_of or std::none_of, sometimes even with std::all_of which improves the readability of the given piece of code a lot.

Go and check in your codebases how std::find_if is used. Do you have similar findings?

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