Class member initialization order, C++

Question

All old and modern C++ books and experts state to initialize class members by their declaration order. But neither explains what if I don't?? I am not talking about classes with members of const types or smth.. just plain simple class.

Consider the sample:

class A
{
  int n;
  std::vector<double> VD;
  char c;
public:
  A():
      VD(std::vector<double>(3)),
      c('a'),
      n(44)
      {
      }
};

Whats the difference of written code and the one with same order in which they are declared???

I don't know about the books, but at least [cppreference.com](http://en.cppreference.com/w/cpp/language/initializer_list#Initialization_order) tells you: >The order of member initializers in the list is *irrelevant*: the actual order of initialization is as follows: ... — Olaf Dietsche, Feb 22 '18 at 14:24

StoryTeller - Unslander Monica · Accepted Answer · 2018-02-22T14:05:31.057

7

Whats the difference of written code and the one with same order in which they are declared???

If members don't depend on each other's initialization order, there is no difference whatsoever. But if they do, then a member initialization list may be telling a lie.

Many a programmer were bitten by this, when they thought their constructors were written correctly, but in fact they had undefined behavior on their hands.

Consider this simple case:

struct foo {
  int _a;
  int _b;
  foo(int b) : _b(b), _a(2 * _b) {}
};

What's _a in the above example? If you answer anything but "the behavior is undefined because _b is used initialized", you'd be wrong.

edited Feb 22 '18 at 14:05

answered Feb 22 '18 at 13:50

StoryTeller - Unslander Monica

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I'm pretty sure that the b being used is the argument not the member - but then this exactly why using the same name for the 2 isn't great. – UKMonkey Feb 22 '18 at 14:04
1

@UKMonkey - Easy bug to re-introduce – StoryTeller - Unslander Monica Feb 22 '18 at 14:05
@UKMonkey - And I didn't even have to use `new`. – StoryTeller - Unslander Monica Feb 22 '18 at 14:07
1

But you could!! `foo(int b) : _b(new int(b)), _a(*b * 2) {}` just working out how to delete it in that line too ... I'm sure it'll involve the comma operator – UKMonkey Feb 22 '18 at 14:09

score 6 · Answer 2 · answered Feb 22 '18 at 13:50

But neither explains what if I don't?

Programmers have no control over it: the order in which you list members in the initialization list has no effect on the actual order of initialization. The compiler ignores the order of items on the list, and re-orders the expressions to match declaration order.

Here is a short example to illustrate this point:

struct Foo {
    Foo(const char *s) { cout << "Init " << s << endl; }
};
struct Bar {
    Foo a;
    Foo b;
    Foo c;
    Bar() : c("c"), b("b"), a("a") {
    }
};

The above prints

Init a
Init b
Init c

even though initialization lists the items in opposite order.

Demo.

score 4 · Answer 3 · answered Feb 22 '18 at 13:47

4

There ought to be absolutely no difference in the generated assembly, although the "as-if" rule might get in the way.

Conceptually at least, n is initialised before c.

Reference: http://en.cppreference.com/w/cpp/language/as_if

answered Feb 22 '18 at 13:47

Bathsheba

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score 3 · Answer 4 · answered Feb 22 '18 at 13:46

You can't change initialisation order - it's the order the members appear in the class - the order in the initialisation list is not significant, though compilers may warn if the two orders don't match up.

score 0 · Answer 5 · answered Feb 22 '18 at 13:51

0

I think there are two reasons to order them properly.

The same order as they are declared makes the code more readable, especially when you want to add or remove some more variables.
The order they are declared indicates their order in the memory. You have more locality when you initialize the variables.

answered Feb 22 '18 at 13:51

Kozuki

97
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On (2) not necessarily if the members are `private`, although I'm not *absolutely* certain of this. A professional like @StoryTeller will hopefully confirm. – Bathsheba Feb 22 '18 at 13:55
@Bathsheba: it is when there are mix of visibility that reordering might happen. – Jarod42 Feb 22 '18 at 13:57
2

All members with the same accessibility are laid out in the order they are declared. But there is no guarantee for members with different accessibility. – StoryTeller - Unslander Monica Feb 22 '18 at 13:57
@StoryTeller: If only reputation were gained by the author on the upvoting of theit comments. – Bathsheba Feb 22 '18 at 14:08
@Bathsheba - We get the "pundit" badge for it. Good enough for me :P – StoryTeller - Unslander Monica Feb 22 '18 at 14:09

score 0 · Answer 6 · answered Feb 22 '18 at 13:59

It depends on how the members are used. If there is a dependency then order must be followed.

Consider below example.

    class x{
         size_t n;
         char * ch; // the size of dynamic char array depends on n
     }

Here, initializing in different order will result in undefined behavior

Other than this reason, of course readability and uniformity matters from coding guidelines POV.

Class member initialization order, C++

6 Answers6