Question

auto a = (Foo<T>*)malloc(sizeof(Foo<T>));
auto *b = (Foo<T>*)malloc(sizeof(Foo<T>));

I don’t think it’s important that templates are there, but the question is: are a and b of the same type?

g++ -std=c++0x -Wall (4.4) doesn’t give any errors or warnings, but I haven’t run the program so I don’t know if it does the same thing.

Does this mean that for a, auto is Foo<T>*, but for b, auto is Foo<T>?

Answer 1

are a and b of the same type?

Let’s find out, shall we?

#include <cstdlib>
#include <type_traits>

template <typename T>
struct Foo
{
    T member;
};

template <typename T>
void test()
{
    auto  a = (Foo<T>*)malloc(sizeof(Foo<T>));
    auto *b = (Foo<T>*)malloc(sizeof(Foo<T>));

    static_assert(std::is_same<decltype(a), decltype(b)>::value, "same type");
}

template void test<int>();   // explicit instantiation

This compiles without a static assertion failure.

Does this mean that for a, auto is Foo<T>*, but for b, auto is Foo<T>?

Yes.