C++ why does empty set::emplace() inserts an element into a set of pointers?

Question

Consider the following code:

struct A{};

int main()
{
    std::set<A*> aset;
    aset.emplace();
    std::cout << aset.size() << std::endl;   //prints "1"
    return 0;
}

DEMO

Why does the empty emplace() adds an element to the set of pointers?

Same reason it'd add an element if the elements weren't pointers. Why would you expect otherwise? (Note that it doesn't construct a pointed-to object.) — user2357112, May 20 '15 at 02:19
@user2357112: but using `std::set aset;` gives a compile-time error. It's not quite the same imo. — davidhigh, May 20 '15 at 02:22
@davidhigh It fails to compile because you have not defined a comparison operator for `A`, not because of your call to `emplace`. (http://coliru.stacked-crooked.com/a/51431ffd8e2b2a1c) — Red Alert, May 20 '15 at 02:25

score 3 · Accepted Answer · answered May 20 '15 at 02:20

Because emplace will:

[Insert] a new element into the container by constructing it in-place with the given args if there is no element with the key in the container.

The container was previously empty, so you're definitely going to insert a new element. Zero arguments is a valid constructor for A*, so the code compiles and you end up with a set with one, value-initialized pointer to A.

C++ why does empty set::emplace() inserts an element into a set of pointers?

1 Answers1