Are std::tuple and std::tuple<std::tuple> considered same type by std::vector?

Question

I have a variable defined like this

auto drum = std::make_tuple
          ( std::make_tuple
          ( 0.3f
          , ExampleClass
          , [](ExampleClass& instance) {return instance.eGetter ();}
          )
          );

I expect drum to be a tuple of tuple. (i.e. ((a, b, c))).

And I have another variable defined like this

auto base = std::make_tuple
          ( 0.48f
          , ExampleClass
          , [](ExampleClass& instance) {return instance.eGetter ();}
          );

which I expect to be just a tuple of three elements (i.e (a, b, c))

I also have a vector defined as follows

std::vector<std::tuple<std::tuple< float
                                 , ExampleClass
                                 , std::function<float (ExampleClass&)>
                                 >>> listOfInstruments;

Now if I add drum to listOfInstruments I expect no errors.

Which was indeed the case with listOfInstruments.push_back(drum);

Where I expected an error was here listOfInstuments.push_back(base); but the code compiles just fine.

Since listOfInstruments has type 'tuple of tuples', should't adding just 'tuple' cause some error? Unless both () and (()) are considered same types by std::vector. Or am I completely wrong and there's something else at work here?

Can't seem to figure it out.

Answer 1

Tuples and vectors are mostly red herrings here. The way that works is simply that push_back, like any function, can perform implicit conversions on its argument, as demonstrated by the following working snippet:

#include <vector>

struct A { };

struct B {
    B(A const &) { }
};

int main() {
    std::vector<B> v;
    v.push_back(A{});
}

Going back to tuple, we can see that it has (among others) a conditionally-explicit constructor (#2 here) which takes references to the tuple's members-to-be:

tuple( const Types&... args );

This constructor is implicit if and only if all of the members have implicit copy constructors, which is the case here (as synthesized constructors are implicit indeed). This means that std::tuple<...> is implicitly convertible to std::tuple<std::tuple<...>>, which is what you're observing.