Does a c++ struct have a default constructor?

The simple answer is yes.
It has a default constructor.

Note: struct and class are identical (apart from the default state of the accesses specifiers).

But whether it initializes the members will depends on how the actual object is declared. In your example no the member is not initialized and a has indeterminate value.

void func()
{
    _bar_  a;                 // Members are NOT initialized.
    _bar_  b = _bar_();       // Members are zero-initialized
    // From C++14
    _bar_  c{};               // New Brace initializer (Members are zero-initialized)


    _bar_* aP = new _bar_;    // Members are NOT initialized.
    _bar_* bP = new _bar_();  // Members are zero-initialized
    // From C++14
    _bar_  cP = new _bar_{};  // New Brace initializer (Members are zero-initialized)
}

// static storage duration objects
//   i.e. objects at the global scope.
_bar_ c; // Members are zero-initialized.

The exact details are explained in the standard at 8.5 Initializers [dcl.init] paragraphs 4-10. But the following is a simplistic summary for this situation.

A structure without a user defined constructor has a compiler generated constructor. But what it does depends on how it is used and it will either default initialize its members (which for POD types is usually nothing) or it may zero initialize its members (which for POD usually means set its members to zero).

PS. Don't use a _ as the first character in a type name. You will bump into problems.

Is it true to say that data members of a struct in c++ are always initialized by default (compared to c)? Or is the observed result just coincidence?

It is a coincidence.

Your code invokes Undefined Behavior; unless you explicitly set the members to 0 they can be anything.

Not an answer, but you might take it to be... if you want to try it:

void foo() {
   struct test {
      int value;
   } x;
   std::cout << x.value << std::endl;
   x.value = 1000;
}
int main() {
   foo();
   foo();
}

In your example, the memory already had the 0 value before the variable was created, so you can call it a lucky coincidence (in fact, some OS will zero out all memory before starting a process, which means that 0 is quite a likely value to find in a small short program...), the previous code will call the function twice, and the memory from the first call will be reused in the second one, chances are that the second time around it will print 1000. Note however that the value is still undefined, and that this test might or not show the expected result (i.e. there are many things that the compiler can do and would generate a different result...)

Member variables of a struct are not initialized by default. Just like a class (because a struct is exactly the same thing as a class, only in struct the members are public by default).

Does a c++ struct have a default constructor?

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