tags:

views:

441

answers:

6
+2  A: 

It's called an initialization list. It initializes members before the body of the constructor executes.

Chris Schmich
+5  A: 

This is called an initialization list. It is for passing arguments to the constructor of a parent class. Here is a good link explaining it: Initialization Lists in C++

dbyrne
+2  A: 

It's called an initialization list. An initializer list is how you pass arguments to your member variables' constructors and for passing arguments to the parent class's constructor.

If you use = to assign in the constructor body, first the default constructor is called, then the assignment operator is called. This is a bit wasteful, and sometimes there's no equivalent assignment operator.

Ken Bloom
+4  A: 

As others have said, it's an initialisation list. You can use it for two things:

  1. Calling base class constructors
  2. Initialising member variables before the body of the constructor executes.

For case #1, I assume you understand inheritance (if that's not the case, let me know in the comments). So you are simply calling the constructor of your base class.

For case #2, the question may be asked: "Why not just initialise it in the body of the constructor?" The importance of the initialisation lists is particularly evident for const members. For instance, take a look at this situation, where I want to initialise m_val based on the constructor parameter:

class Demo
{
    Demo(int& val) 
     {
         m_val = val;
     }
private:
    const int& m_val;
};

By the C++ standard, this is illegal. We cannot change the value of a const variable in the constructor, because it is marked as const. So you can use the initialisation list:

class Demo
{
    Demo(int& val) : m_val(val)
     {
     }
private:
    const int& m_val;
};

That is the only time that you can change a const member variable. And as Michael noted in the comments section, it is also the only way to initialise a reference that is a class member.

Outside of using it to initialise const member variables, it seems to have been generally accepted as "the way" of initialising variables, so it's clear to other programmers reading your code.

Smashery
To be clear - in addition to initializing `const` members, it's the only way to initialize reference members, whether `const` or not (since an assignment to a reference is an assignment to what the reference refers to).
Michael Burr
+1  A: 

You are calling the constructor of its base class, demo.

Chris O
+1  A: 

It means that len is not set using the default constructor. while the demo class is being constructed. For instance:

class Demo{
    int foo;
public:
    Demo(){ foo = 1;}
};

Would first place a value in foo before setting it to 1. It's slightly faster and more efficient.

wheaties