Why does the standard library have find and find_if?

感情迁移 提交于 2019-12-01 14:19:42

问题


Couldn't find_if just be an overload of find? That's how std::binary_search and friends do it...


回答1:


A predicate is a valid thing to find, so you could arrive at ambiguities.


Consider find_if is renamed find, then you have:

template <typename InputIterator, typename T>
InputIterator find(InputIterator first, InputIterator last, const T& value);

template <typename InputIterator, typename Predicate>
InputIterator find(InputIterator first, InputIterator last, Predicate pred);

What shall be done, then, with:

find(c.begin(), c.end(), x); // am I finding x, or using x to find?

Rather than try to come up with some convoluted solution to differentiate based on x (which can't always be done*), it's easier just to separate them.

*This would be ambiguous, no matter what your scheme is or how powerful it might be†:

struct foo
{
    template <typename T>
    bool operator()(const T&);
};

bool operator==(const foo&, const foo&);

std::vector<foo> v = /* ... */;
foo f = /* ... */; 

// f can be used both as a value and as a predicate
find(v.begin(), v.end(), f); 

†Save mind reading.




回答2:


Here's what Stroustrup said (The C++ Programming Language, 18.5.2):

If find() and find_if() had the same name, surprising abmiguities would have resulted. In general, the _if suffix is used to indicate that an algrithm takes a predicate.

As to what exactly that "ambiguity" is, Steve Jessop answered that in his (top rated) answer to this SO question.

(note: that question may actually qualify as the same question as this one. I'm not quite smart enough in C++ arcania to decide).




回答3:


It can't have the same name because there would be an ambiguity. Suppose that we had a find overload instead of find_if. Then suppose:

// Pseudo-code
struct finder
{
    bool operator()(const T&) const { ... }
    bool operator==(const finder& right) const { ... }
}

std::vector<finder> finders;

finder my_finder;

std::find(finders.begin(), finders.end(), my_finder);

The find would have no way to resolve the inconsistency: Should it attempt to find the finder in the container, or use the finder to do the find operation? To solve this problem they created two function names.




回答4:


You can certainly implement find in terms of find_if using some sort of equality predicate.

I would guess that the real reason is that you can implement find fairly easily and provide performant specialised implementations for typical encountered types; if you are using find_if, the predicate you pass in can be arbitrarily complex, which gives the library implementer less scope of optimisation.

Also, C++ has the philosphy of "you don't pay for what you don't use" and you'd normally expect that you don't want to pay for a predicate evaluation if a simple comparison will do.



来源:https://stackoverflow.com/questions/3533251/why-does-the-standard-library-have-find-and-find-if

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