Proper way of transferring ownership of a std::vector< std::unique_ptr< int> > to a class being constructed

std::unique_ptr<T> is a non-copyable but movable type. Having a move-only type in a std:vector<T> make the std::vector<T> move-only, too. To have the compiler automatically move objects, you need to have an r-value for move-construction or move-assignment. Within your constructor the object vecOfIntPtrsOwnedByCaller is an l-value, although one which, despite its name, already owns the pointed to ints: they got “stolen” from the caller when the caller created the object. To move from an l-value, you need to use std::move() (or something equivalent):

Foo::Foo(std::vector<std::unique_ptr<int>> vecOfIntPtrsOwnedByCaller)
{
    _vecOfIntPtrsOwnedByFoo = std::move(vecOfIntPtrsOwnedByCaller);
}

or, preferable

Foo::Foo(std::vector<std::unique_ptr<int>> vecOfIntPtrsOwnedByCaller)
    : _vecOfIntPtrsOwnedByFoo(std::move(vecOfIntPtrsOwnedByCaller))
{
}

The latter approach avoid first default-constructing the member and then move-assigning to it and, instead, move-constructs the member directly. I guess, I would also make the argument an r-value reference but this isn’t necessary.

Note, that you can construct objects of type Foo only from something which can be bound to an r-value, e.g.:

int main() {
    Foo f0(std::vector<std::unique_ptr<int>>()); // OK
    std::vector<std::unique_ptr<int>> v;
    Foo f1(v); v// ERROR: using with an l-value
    Foo f2{v}; v// ERROR: using with an l-value
    Foo f3 = v; // ERROR: using with an l-value
    Foo f4(std::move(v)); // OK: pretend that v is an r-value
}