Mehrdad Mehrdad - 2 months ago 8
C++ Question

Does a default constructor always initialize all members?

I could swear I don't remember having seen this before, and I'm having trouble believing my eyes:

Does an implicitly-defined default constructor for a non-aggregate class initialize its members or no?

In Visual C++, when I run this innocent-looking code...

#include <string>
struct S { int a; std::string b; };
int main() { return S().a; }


... to my astonishment, it returns a non-zero value! But if I remove field
b
, then it returns zero.

I've tried this on all versions of VC++ I can get my hands on, and it seems to do this on all of them.

But when I try it on Clang and GCC, the values are initialized to zero, whether I try it in C++98 mode or C++11 mode.

What's the correct behavior? Is it not guaranteed to be zero?

hvd hvd
Answer

Quoting C++11:

5.2.3 Explicit type conversion (functional notation) [expr.type.conv]

2 The expression T(), where T is a simple-type-specifier or typename-specifier for a non-array complete object type or the (possibly cv-qualified) void type, creates a prvalue of the specified type,which is value-initialized (8.5; no initialization is done for the void() case). [...]

8.5 Initializers [dcl.init]

7 To value-initialize an object of type T means:

  • ...
  • if T is a (possibly cv-qualified) non-union class type without a user-provided constructor, then the object is zero-initialized and, if T's implicitly-declared default constructor is non-trivial, that constructor is called.
  • ...

So in C++11, S().a should be zero: the object is zero-initialized before the constructor gets called, and the constructor never changes the value of a to anything else.

Prior to C++11, value initialization had a different description. Quoting N1577 (roughly C++03):

To value-initialize an object of type T means:

  • ...
  • if T is a non-union class type without a user-declared constructor, then every non-static data member and base-class component of T is value-initialized;
  • ...
  • otherwise, the object is zero-initialized

Here, value initialization of S did not call any constructor, but caused value initialization of its a and b members. Value initialization of that a member, then, caused zero initialization of that specific member. In C++03, the result was also guaranteed to be zero.

Even earlier than that, going to the very first standard, C++98:

The expression T(), where T is a simple-type-specifier (7.1.5.2) for a non-array complete object type or the (possibly cv-qualified) void type, creates an rvalue of the specified type, whose value is determined by default-initialization (8.5; no initialization is done for the void() case).

To default-initialize an object of type T means:

  • if T is a non-POD class type (clause 9), the default constructor for T is called (and the initialization is ill-formed if T has no accessible default constructor);
  • ...
  • otherwise, the storage for the object is zero-initialized.

So based on that very first standard, VC++ is correct: when you add a std::string member, S becomes a non-POD type, and non-POD types don't get zero initialization, they just have their constructor called. The implicitly generated default constructor for S does not initialise the a member.

So all compilers can be said to be correct, just following different versions of the standard.

As reported by @Columbo in the comments, later versions of VC++ do cause the a member to be initialized, in accordance with more recent versions of the C++ standard.

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