Is the order of initialization guaranteed by the standard?

Question

In the following code snippet, d1's initializer is passed d2 (which has not been constructed yet, correct?)

So is the d.j in D's copy constructor an uninitialized memory access?

struct D
{
    int j;

    D(const D& d) { j = d.j; }
    D(int i) { j = i; }
};

struct A
{
    D d1;
    D d2;
    A() : d2(2), d1(d2) {}
};

Which section of the C++ standard discusses the initialization order of data members?

Related though not duplicate question: http://stackoverflow.com/questions/1589950/initializer-list-argument-evaluation-order — Johannes Schaub - litb, Nov 21 '09 at 05:32

score 12 · Accepted Answer · answered Nov 21 '09 at 01:33

12

I don't have the standard handy right now so I can't quote the section, but structure or class member initialisation always happens in declared order. The order in which members are mentioned in the constructor initialiser list is not relevant.

Gcc has a warning -Wreorder that warns when the order is different:

       -Wreorder (C++ only)
           Warn when the order of member initializers given in the code does
           not match the order in which they must be executed.  For instance:

                   struct A {
                     int i;
                     int j;
                     A(): j (0), i (1) { }
                   };

           The compiler will rearrange the member initializers for i and j to
           match the declaration order of the members, emitting a warning to
           that effect.  This warning is enabled by -Wall.

answered Nov 21 '09 at 01:33

Greg Hewgill

951,095
183
1,149
1,285

5

The reason for this is that things are destructed in the reverse order they are constructed. You can't change the order members get destructed after the destructor, therefore the constructed order *must* correspond. – GManNickG Nov 21 '09 at 01:36
2

You can however, still refer to previously initialized members, as discussed here: http://www.parashift.com/c++-faq-lite/ctors.html#faq-10.7 – Jherico Nov 21 '09 at 01:37
1

The FAQ explanation in this case is pretty useful, I think; the standard says "there is a sequence point after the initialization of each base and member" (12.6.2/3). – James McNellis Nov 21 '09 at 02:54

score 12 · Answer 2 · answered Nov 21 '09 at 01:41

The C++ standard (ISO/IEC 14882:2003 12.6.2/5, Initializing bases and members) says:

Initialization shall proceed in the following order:

— First, and only for the constructor of the most derived class as described below, virtual base classes shall be initialized in the order they appear on a depth-first left-to-right traversal of the directed acyclic graph of base classes, where “left-to-right” is the order of appearance of the base class names in the derived class base-specifier-list.

— Then, direct base classes shall be initialized in declaration order as they appear in the base-specifier-list (regardless of the order of the mem-initializers).

— Then, nonstatic data members shall be initialized in the order they were declared in the class definition (again regardless of the order of the mem-initializers).

— Finally, the body of the constructor is executed.

Bullet point 3 guarantees the order of nonstatic data member initialization.

score 6 · Answer 3 · answered Nov 21 '09 at 01:37

In your example it will fail:

struct A
{
    D d1, d2;    
    A() : d2(2), d1(d2) {}
};

d1: is initialised first as it is declared first.
d2: is then initialized.

As a result the initializer list will construct d1 using a reference to an invalid object (d2).

This is one reason to turn up your compilers warning level as high as possable.
And additionaly force it to report all warnings as errors.

score 2 · Answer 4 · answered Nov 21 '09 at 01:39

This phenomenon is explained/highlighted in Item 13 of Meyer's Effective C++. It says that the destructor must destroy elements in the inverse order of its constructors, therefore all constructors must initialize elements in the same order, therefore they initialize them in the order in which they're declared (instead of the sequence of the initialization lists).

score 1 · Answer 5 · answered Nov 21 '09 at 01:35

1

Yes. A good compiler should warn you that A::d2 will be initialized after A::d1.

answered Nov 21 '09 at 01:35

wilhelmtell

57,473
20
96
131

Is the order of initialization guaranteed by the standard?

5 Answers5

Linked

Related