Why differentiate predicate and non-predicate versions for generic algorithms?

Question

The standard library does differentiate predicate and non-predicate versions of generic algorithms. For example, std::sort() looks like:

template< class RandomIt >
void sort( RandomIt first, RandomIt last );

template< class RandomIt, class Compare >
void sort( RandomIt first, RandomIt last, Compare comp );

Is there any problem in just writing the following?

template< class RandomIt, class Compare = std::less<void>>
void sort( RandomIt first, RandomIt last, Compare comp = Compare{});

There was no such thing as default template arguments for function templates in C++03. — T.C., Jan 28 '16 at 08:30
@T.C. So, suppose we just redesign the whole thing, should the second form be preferred? — Lingxi, Jan 28 '16 at 08:32
@T.C. I believe the default template argument thing is actually an answer. — Angew is no longer proud of SO, Jan 28 '16 at 08:33
@BoBTFish I think `std::less` does just that (`operator <`). No? — Lingxi, Jan 28 '16 at 08:35
@Lingxi You can specialize `std::less` for your types and make it do something other. — Revolver_Ocelot, Jan 28 '16 at 08:37

score 8 · Accepted Answer · answered Jan 28 '16 at 08:40

Pretty much historical reasons.

C++98/03 didn't have default template arguments for function templates, so it had to use two overloads. And changing it afterwards can break user code.

So, suppose we just redesign the whole thing, should the second form be preferred?

That's what the current Ranges TS working draft does.

Why differentiate predicate and non-predicate versions for generic algorithms?

1 Answers1