Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

What's the differences between member initializer list and default member initializer on non-static data member?

I'd like to understand what's the differences of using one form rather than the other (if any).

Code 1 (init directly on variables):

#include <iostream>

using namespace std;

class Test 
{
public:
    Test() {
        cout<< count;
    }

    ~Test();

private:
    int count=10;
};

int main()
{
    Test* test = new Test();
}

Code 2 (init with initialization list on constructor):

#include <iostream>

using namespace std;

class Test 
{
public:
    Test() : count(10) {
        cout<< count;
    }

    ~Test();

private:
    int count;
};

int main()
{
    Test* test = new Test();
}

Is there any difference in the semantics, or it is just syntactic?

like image 304
markzzz Avatar asked Apr 13 '16 13:04

markzzz


2 Answers

Member initialization

In both cases we are talking about member initialization. Keep in mind that the members are initialized in the sequence in which they are declared in the class.

Code 2: Member initializer list

In the second version:

Test() : count(10) {

: count(10) is a constructor initializer (ctor-initializer) and count(10) is a member initializer as part of the member initializer list. I like to think of this as the 'real' or primary way that the initialization happens, but it does not determine the sequence of initialization.

Code 1: Default member initializer

In the first version:

private:
    int count=10;

count has a default member intitializer. It is the fallback option. It will be used as a member initializer if none is present in the constructor, but in the class the sequence of members for initialization is determined.

From section 12.6.2 Initializing bases and members, item 10 of the standard:

If a given non-static data member has both a brace-or-equal-initializer and a mem-initializer, the initialization specified by the mem-initializer is performed, and the non-static data member’s brace-or-equal-initializer is ignored. [ Example: Given

struct A {
int i = / some integer expression with side effects / ;
A(int arg) : i(arg) { }
// ...
};

the A(int) constructor will simply initialize i to the value of arg, and the side effects in i’s brace-or-equalinitializer will not take place. —end example ]

Something else to keep in mind would be that if you introduce a non-static data member initializer then a struct will no longer be considered an aggregate in C++11, but this has been updated for C++14.


Differences

what's the differences of using one form rather than the other (if any).

  • The difference is the priority given to the two options. A constructor initializer, directly specified, has precedence. In both cases we end up with a member initializer via different paths.
  • It is best to use the default member initializer because
    • then the compiler can use that information to generate the constructor's initializer list for you and it might be able to optimize.
    • You can see all the defaults in one place and in sequence.
    • It reduces duplication. You could then only put the exceptions in the manually specified member initializer list.
like image 54
wally Avatar answered Oct 17 '22 02:10

wally


In the C++ Core Guidelines (see note 1 below), Guideline C.48 recommends the first approach (in-class initializers.) The reasoning provided is:

Makes it explicit that the same value is expected to be used in all constructors. Avoids repetition. Avoids maintenance problems. It leads to the shortest and most efficient code.

In fact if your constructor does nothing but initialize member variables, as in your question, then Guideline C.45 is firmer still, saying to use in-class initializers for sure. It explains that

Using in-class member initializers lets the compiler generate the function for you. The compiler-generated function can be more efficient.

I am not going to argue with Stroustrup, Sutter, and several hundred of their friends and colleagues even if I haven't written a compiler so I can't prove it's more efficient. Use in-class initializers wherever you can.

  1. If you're not familiar with the guidelines do follow the links to see sample code and more explanations.
like image 12
Kate Gregory Avatar answered Oct 17 '22 02:10

Kate Gregory