why c++ use memset(addr,0,sizeof(T)) to construct a object? Standard or compiler bug?

Question

This question is related to another post of mine: why allocate_shared and make_shared so slow

In here I can describe the question more clearly.

Think about t

Answer 1

A a(t...);

Will be parsed as initializing a with t....^† When t... is empty, as when you call it, this will be understood as value-initializing a.

For A without a user-provided default constructor, value-initialize is to zero all its members, hence the memset.

When you provide a constructor for A, value-initialize is to call the default constructor, which you defined to be do nothing, therefore no memset will be called.

This is not a bug in the compiler, this is required behaviour. To remove the redundant memset, you could just write A a;. In this case a is default-initialized and no automatic zeroing occurs, with or without the user-provided constructor.

_{† This is important since A a() will be parsed as a function called a with return type A}

Answer 2

Doesn't this explain it?

We can see that:

Zero initialization is performed [...] as part of value-initialization sequence for [...] members of value-initialized class types that have no constructors, including value initialization of elements of aggregates for which no initializers are provided.

...

Value initialization is performed [...] when a non-static data member or a base class is initialized using a member initializer with an empty pair of parentheses or braces (since C++11);

So putting a_() in the member initializer list falls into the latter case, that as a result invokes zero initialization of the array.

To answer your question: to me this seems to be a standard behavior not a compiler bug.