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What is this weird colon-member (" : ") syntax in the constructor?

Recently I've seen an example like the following:

#include <iostream>

class Foo {
public:
  int bar;
  Foo(int num): bar(num) {};
};

int main(void) {
  std::cout << Foo(42).bar << std::endl;
  return 0;
}

What does this strange : bar(num) mean? It somehow seems to initialize the member variable but I've never seen this syntax before. It looks like a function/constructor call but for an int? Makes no sense for me. Perhaps someone could enlighten me. And, by the way, are there any other esoteric language features like this, you'll never find in an ordinary C++ book?

An "ordinary c++ book" that doesn't mention this is probably a c book where someone thought "++" would look cool on the cover ...
"you'll never find in a ordinary C++ book". Oh. Dear. Throw away your "ordinary C++ book" right now. Not out the window - somebody else might pick it up. Preferably shred it and put it for recycling. Done? Now consult stackoverflow.com/questions/388242/… to get a new book.
This language feature is hardly esoteric. It's a fairly major feature of object construction.
In fact, far from esoteric, you often have no choice but to use initializer lists. For example, if your class contains a const member variable, or a reference, you have to use an initalizer list.

o
orestisf
Foo(int num): bar(num)    

This construct is called a Member Initializer List in C++.

Simply said, it initializes your member bar to a value num.

What is the difference between Initializing and Assignment inside a constructor?

Member Initialization:

Foo(int num): bar(num) {};

Member Assignment:

Foo(int num)
{
   bar = num;
}

There is a significant difference between Initializing a member using Member initializer list and assigning it an value inside the constructor body.

When you initialize fields via Member initializer list the constructors will be called once and the object will be constructed and initialized in one operation.

If you use assignment then the fields will be first initialized with default constructors and then reassigned (via assignment operator) with actual values.

As you see there is an additional overhead of creation & assignment in the latter, which might be considerable for user defined classes.

Cost of Member Initialization = Object Construction 
Cost of Member Assignment = Object Construction + Assignment

The latter is actually equivalent to:

Foo(int num) : bar() {bar = num;}

While the former is equivalent to just:

Foo(int num): bar(num){}

For an inbuilt (your code example) or POD class members there is no practical overhead.

When do you HAVE TO use Member Initializer list?

You will have(rather forced) to use a Member Initializer list if:

Your class has a reference member

Your class has a non static const member or

Your class member doesn't have a default constructor or

For initialization of base class members or

When constructor’s parameter name is same as data member(this is not really a MUST)

A code example:

class MyClass {
public:
  // Reference member, has to be Initialized in Member Initializer List
  int &i;
  int b;
  // Non static const member, must be Initialized in Member Initializer List
  const int k;

  // Constructor’s parameter name b is same as class data member
  // Other way is to use this->b to refer to data member
  MyClass(int a, int b, int c) : i(a), b(b), k(c) {
    // Without Member Initializer
    // this->b = b;
  }
};

class MyClass2 : public MyClass {
public:
  int p;
  int q;
  MyClass2(int x, int y, int z, int l, int m) : MyClass(x, y, z), p(l), q(m) {}
};

int main() {
  int x = 10;
  int y = 20;
  int z = 30;
  MyClass obj(x, y, z);

  int l = 40;
  int m = 50;
  MyClass2 obj2(x, y, z, l, m);

  return 0;
}

MyClass2 doesn't have a default constructor so it has to be initialized through member initializer list.

Base class MyClass does not have a default constructor, So to initialize its member one will need to use Member Initializer List.

Important points to Note while using Member Initializer Lists:

Class Member variables are always initialized in the order in which they are declared in the class.

They are not initialized in the order in which they are specified in the Member Initializer List. In short, Member initialization list does not determine the order of initialization.

Given the above it is always a good practice to maintain the same order of members for Member initialization as the order in which they are declared in the class definition. This is because compilers do not warn if the two orders are different but a relatively new user might confuse member Initializer list as the order of initialization and write some code dependent on that.


@nils This is the best answer so far. The order of the initialization Als pointed out is also extremely important, while the Visual Studio compiler won't say anything about that, other compiler like gcc will fail. It is also important to note depending your compiler and the situation it is not Always true that this will improve performance or be more efficient.
@ryf9059: Why do you think it will be inconvenient? You have to list them anyways, so why not in order same as that of declaration.
this should have been the answer. thank god i scrolled down else i would have missed it.
@AlokSave MyClass(int a, int b, int c) : i(a), b(b), k(c) { // Without Member Initializer // this->b = b; } it should be like this MyClass(int &a , int b, int c) : i(a), b(b), k(c) { // Without Member Initializer // this->b = b; } and so respective changes in declaration and call. without this change i will refer to a but a can't refer to x as it contain just value of x so indirectly i can't refer to x. so if we modify value of i then it just modify a but not x
@AbhishekMane You are right, and here is a link to the related question showing it: stackoverflow.com/q/67619383/3150802
T
TallChuck

It's a member initialization list. You should find information about it in any good C++ book.

You should, in most cases, initialize all member objects in the member initialization list (however, do note the exceptions listed at the end of the FAQ entry).

The takeaway point from the FAQ entry is that,

All other things being equal, your code will run faster if you use initialization lists rather than assignment.


Knowing the terminology is critical - I'm jealous that I didn't think of it.
There are also a bunch of other reasons for using init lists. especially when the order of initialisation matters. It's a shame it has such a stupid fake function call syntax.
@mgb, the initialization list does not determine the order of initialization. Member variables are initialized in the order they are declared in the class, even if that differs from the order of the initializations on the constructor.
@mgb: I don't think it's intended as a fake function call syntax. It's initialization syntax, like int i(23);, std::vector<double> emptyVec(0);, std::vector<double> fullVec(10,23.);, etc. Only with the type removed, of course, because the type is in the member declaration.
@Martin : It doesn't have a function call syntax, it has a construction syntax (ala: new String("Name")). It fits in with the constructor better than Foo(int num) : m_Count = 5. Not to mention that classes must be constructed at this point anyway, since it's initialized here. Foo(int num) : Bar = num, wouldn't compile right. It just seems weird seeing Foo(int num) : m_Count(num), since primitive types aren't constructed.
L
Lightness Races in Orbit

That's constructor initialisation. It is the correct way to initialise members in a class constructor, as it prevents the default constructor being invoked.

Consider these two examples:

// Example 1
Foo(Bar b)
{
   bar = b;
}

// Example 2
Foo(Bar b)
   : bar(b)
{
}

In example 1:

Bar bar;  // default constructor
bar = b;  // assignment

In example 2:

Bar bar(b) // copy constructor

It's all about efficiency.


I wouldn't say that this is about efficiency. It is about providing a way to initialize something that requires initialization, but cannot be default-initualized. For some reason people mention constants and references as the examples, while the most obvious example would be classes without default constructors.
We're both right; in his example you could argue for efficiency; for the const/reference/no default constructor issue it's both efficiency and necessity. I upvoted an answer below because of it :) [Farnsworth voice] It can do other things. Why shouldn't it?
Bar bar(); // default constructor Are you sure?
@LightnessRacesinOrbit Just curious to know : What should that be according to you?
L
LeopardSkinPillBoxHat

This is called an initialization list. It is a way of initializing class members. There are benefits to using this instead of simply assigning new values to the members in the body of the constructor, but if you have class members which are constants or references they must be initialized.


@LightnessRacesinOrbit - Point taken, but the constants or references point is still valid in my answer.
w
wkl

This is not obscure, it's the C++ initialization list syntax

Basically, in your case, x will be initialized with _x, y with _y, z with _z.


D
Destructor

The other already explained to you that the syntax that you observe is called "constructor initializer list". This syntax lets you to custom-initialize base subobjects and member subobjects of the class (as opposed to allowing them to default-initialize or to remain uninitialized).

I just want to note that the syntax that, as you said, "looks like a constructor call", is not necessarily a constructor call. In C++ language the () syntax is just one standard form of initialization syntax. It is interpreted differently for different types. For class types with user-defined constructor it means one thing (it is indeed a constructor call), for class types without user-defined constructor it means another thing (so called value initialization ) for empty ()) and for non-class types it again means something different (since non-class types have no constructors).

In your case the data member has type int. int is not a class type, so it has no constructor. For type int this syntax means simply "initialize bar with the value of num" and that's it. It is done just like that, directly, no constructors involved, since, once again, int is not a class type of therefore it can't have any constructors.


But bjarne stroustrup in his book TC++PL & The C++ Programming language says that "Built-in types also have default constructors" . geeksforgeeks.org/c-default-constructor-built-in-types & informit.com/guides/content.aspx?g=cplusplus&seqNum=15 also says built in types have constructors. I personally asked this question to bjarne via mail & he says me that yes, built in types also have constructors. so your answer is wrong !!!
@Destructor: My answer is absolutely correct. Bjarne Stroustrup deliberately and explicitly lied in his book for the sake of simplifying it. Compare the size of the TC++PL book and the size of C++ language standard. See the difference? The price to pay for relative compactness of TC++PL is such obvious (and well-known to everyone) errors and omissions as the one you mentioned (there are quite a few others as well). So, to put it more succinctly: my answer is right, TC++PL is wrong. But for a person like you, who is just beginning to learn, TC++PL is good enough.
And please don't tell us trollish fairy tales about "asked this question to bjarne via mail". This issue is, again, well-known and has been exhaustively discussed and closed a long time ago. (But even if he said something like that to you, it wouldn't matter anyway.) In the end, the only thing that matters is what the language specification says. What Bjarne Stroustrup says is irrelevant. The post on geeksforgeeks you linked is completely bogus.
So, you think that standard is bug free? C++ standard has also bugs.
@Destructor: Of course, it does. However, the standard defines the language. Everything the standard says is The Absolute Truth by definition. The only "bugs" it can possibly have are mostly wording-related things like self-contradictory wording, ambiguous wording, under-specified and so forth. Such bugs are vehemently sought out, reported, documented, discussed and resolved. "Bugs" in the intent can also exist, but they are a matter of debate.
M
Mark Ransom

I don't know how you could miss this one, it's pretty basic. That's the syntax for initializing member variables or base class constructors. It works for plain old data types as well as class objects.


Written on one line inside the declaration like that it's easy to not spot it as an init list
n
nos

This is an initialization list. It'll initialize the members before the constructor body is run. Consider

class Foo {
 public:
   string str;
   Foo(string &p)
   {
      str = p;
   };
 };

vs

class Foo {
public:
  string str;
  Foo(string &p): str(p) {};
};

In the first example, str will be initialized by its no-argument constructor

string();

before the body of the Foo constructor. Inside the foo constructor, the

string& operator=( const string& s );

will be called on 'str' as you do str = p;

Wheras in the second example, str will be initialized directly by calling its constructor

string( const string& s );

with 'p' as an argument.


A
Aric TenEyck

You are correct, this is indeed a way to initialize member variables. I'm not sure that there's much benefit to this, other than clearly expressing that it's an initialization. Having a "bar=num" inside the code could get moved around, deleted, or misinterpreted much more easily.


The benefit is that it's usually more efficient. And, in some cases, such as when you have const member variables, or member variables that are references, you have to use an initializer list.
p
pm100

there is another 'benefit'

if the member variable type does not support null initialization or if its a reference (which cannot be null initialized) then you have no choice but to supply an initialization list


s
suszterpatt

It's an initialization list for the constructor. Instead of default constructing x, y and z and then assigning them the values received in the parameters, those members will be initialized with those values right off the bat. This may not seem terribly useful for floats, but it can be quite a timesaver with custom classes that are expensive to construct.


M
M.M

Not mentioned yet on this thread: since C++11, the member initializer list can use list-initialization (aka. "uniform initialization", "braced initialization"):

Foo(int num): bar{num} {}

which has the same semantics as list-initialization in other contexts.


K
Karen Baghdasaryan

Although this is an old discussion, I couldn't find any mention about delegating constructor, which uses the weird ":" symbol in the following way.

class Foo 
{
public: 
    Foo(char x, int y) 
    {}
    Foo(int y) : Foo('a', y) 
    {}
};

What it does is simply delegating Foo(y) into Foo('a', y) . So that

Foo foo(15); //=> foo('a', 15)

When defining a delegating constructor, you cannot have any members in initializer list besides targeted constructor.