Explanation of C++ class constructor syntax? [duplicate] - c++

This question already has answers here:
What is this weird colon-member (" : ") syntax in the constructor?
(14 answers)
Closed 5 years ago.
Reading through The C++ Programming Language, 4th Edition, there's a class defined like so
class Vector
{
private:
int sz;
double *a;
public:
Vector(int s) :elem{new double[s]}, sz{s} {}
}
I'm a bit confused on how this constructor syntax works. I believe Vector(int s) is creating a constructor function that takes one parameter s, and that it initializes elem and sz. But why is there a :? I thought functions bodies were surrounded by {}? And so what do the empty braces {} at the end serve?

: is called an initialiser list, which is used to quickly and concisely set values for the member variables when the constructor is called.
{} is the constructor's method body. Since the constructor is similar to a method, there has to be a body present for the code to compile. Since there is no need for any code in there, an empty body is used so the function does nothing.

This is initialization with Initializer List.

: is used to "initialize" the members of a class (this method is also called
member initialization list)
there is a major difference between using : and function body {}
initiallizer list : initialize the members of class, whereas ,constructor body {} assigns the value to the members of the class.
the difference may not seem very big but it is actually the only way to initialize the const data type and reference data type members (which can only be initialized during declaration )
So when you do this
class Test
{
const int i; const string str;
public:
Test(int x, string y):i{x},str{y};
}
This would work, but if you try to assign values to const int i and const string str by writing their code in the body of constructor, it would lead to a result
And so what do the empty braces {} at the end serve?
nothing it is just compulsory to put those braces (even if it is empty)
They can basically serve as a function when you create an object of the class inside the main function and pass it the required arguments.

Related

C++ Xcode Error When Initializing a Class With Inputs Within Another Class’ Header [duplicate]

This question already has answers here:
Why can't member initializers use parentheses?
(2 answers)
Closed 7 months ago.
I have a line of code inside a class's private member variables:
vector<double> dQdt(3)
When compiling in xcode, this gives an error "expected parameter declarator." I think I provided sufficient info. I don't see anything wrong with this declaration.
You have to initialize the variable in the constructor's initializer list:
class X
{
private:
vector<double> dQdt;
public:
X() : dQdt(3) {}
};
If you read e.g. this member initialization reference you will learn that default member initialization have to be a brace or equals initializer. I.e. you need to either use curly-braces:
std::vector<double> dQdt{ 0.0, 0.0, 0.0 };
or using the equals character:
std::vector<double> dQdt = std::vector<double>(3);
Since this was introduced with the C++11 standard, you need to enable that in your environment.
I got this error while trying to compile my C++ code having an initialized vector.
Change the () to {} worked for me in the initialization part;
Earlier my code looked like this:
vector<vector<int>> minA(11, vector<int>(11, INT_MAX));
I changed my code to replace circular brackets with curly braces and the error disappered.
vector<vector<int>> minA{11, vector<int>(11, INT_MAX)};
The parameter for constructors of data members should be written in the initializer list of your class' constructor. That is, instead of
class Foo {
vector<double> dQdt(3);
};
You should write
class Foo {
public:
Foo() : dQdt(3) {}
private:
vector<double> dQdt;
};
As well as initializing in the initializer list of the constructor, you can initialize with a brace initializer list:
class Foo {
vector<double> dQdt{3};
};
The actual text of the error is because the compiler was expecting you to declare a function, taking an argument of some type, and return the vector<double>. 3 is not a valid declaration of a parameter to a function.
The question is already answered the following however works as well.
( Which might be more useful to assign initial values. For example 24 times the 42. )
const int default_value = 42;
struct foo
{
vector<double> hour{vector<double>(24,default_value)};
};

Defining Function in c++ in different style [duplicate]

This question already has answers here:
Closed 12 years ago.
Possible Duplicates:
Variables After the Colon in a Constructor
C++ constructor syntax question (noob)
I have some C++ code here:
class demo
{
private:
unsigned char len, *dat;
public:
demo(unsigned char le = 5, unsigned char default) : len(le)
{
dat = new char[len];
for (int i = 0; i <= le; i++)
dat[i] = default;
}
void ~demo(void)
{
delete [] *dat;
}
};
class newdemo : public demo
{
private:
int *dat1;
public:
newdemo(void) : demo(0, 0)
{
*dat1 = 0;
return 0;
}
};
My question is, what are the : len(le) and : demo(0, 0) called?
Is it something to do with inheritance?
As others have said, it's an initialisation list. You can use it for two things:
Calling base class constructors
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++ specification, 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.
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++
It's called an initialization list. It initializes members before the body of the constructor executes.
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.
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.
You are calling the constructor of its base class, demo.

Why do you have to define an empty constructor c++ [duplicate]

This question already has answers here:
When do we need to have a default constructor?
(7 answers)
Closed 8 years ago.
I have the following classes:
class ArithmeticExpression
{
public:
ArithmeticExpression(std::string expr);
}
class Command
{
public:
Command(){};
//this is a virtual class
}
class CommandAssign : public Command
{
private:
ArithmeticExpression expr;
public:
CommandAssign();
CommandAssign(ArithmeticExpression);
}
Now when I try to write the constructor for the CommandAssign class as in:
CommandAssign::CommandAssign(ArithmeticExpression expr)
:Command()
{
this -> expr = ArithmeticExpression(expr.getExpr());
}
I get the error:
no matching function for call to ‘ArithmeticExpression::ArithmeticExpression()’
:Command()
Apparently I can fix that by adding an empty constructor in ArithmeticExpression class that does not do anything. What is it so special about this empty constructor that makes it work? I do not explicitly call anywhere. Do you always need to define an empty constructor in C++?
I wanted to emphasize that although from the title it seems that my question is similar to the one some users suggested as being a duplicate of, the answer I was looking for is NOT there. I was simply trying to understand what happens when a constructor is called and how to avoid defining a useless default constructor, which I knew already is not automatically defined by the compiler in the case where I define one with parameters.
A default constructor will only be automatically generated by the compiler if no other constructors are defined.
EDIT:
The default constructor is needed for object initialization.
All members are initialised before the constructor body begins. If one doesn't have an entry in the initialiser list, then it will be default-initialised; but this is only possible (for a class type) if it has a default constructor.
expr is not initialised in the initialiser list, and doesn't have a default constructor (since declaring any constructor prevents one from being implicitly generated), so it can't be initialised - hence the error.
You should initialise it in the list, rather than reassigning it in the constructor body:
CommandAssign::CommandAssign(ArithmeticExpression expr) :
expr(expr.getExpr())
{}
Note that there's no need to explicitly default-construct the Command sub-object. This also requires the constructor of ArithmeticExpression to be public: it's private in your example code.

Initialization of Members in constructor [duplicate]

This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
C++ initialization lists
class Base
{
public:
int m_nValue;
Base(int nValue=0)
: m_nValue(nValue)
{
}
};
In this code, is the constructor initializing m_nvalue member variable?
I am not sure of this syntax:
Base(int nValue=0) : m_nValue(nValue) {}
We normally write it as:
Base(int nValue) { m_nValue = nValue;}
Can some one explain the above syntax of C++?
This syntax:
Base(int nValue=0)
: m_nValue(nValue)
is called the member initializer. It will initialize m_nValue with given nValue. This syntax is usually preferred in C++ since it is executed before the body of the constructor.
It's called member initializer list.
The member initializer list consists of a comma-separated list of initializers preceded by a colon. It’s placed after the closing
parenthesis of the argument list and before the opening bracket of the function body
Conceptually, these initializations
take place when the object is created and before any code within the brackets is executed.
Note:
You can’t use the member initializer list syntax with class methods other than constructors.
The way of initializing a variable in your code is called as member initializer list.
Generally we use such list to initialize const member variable (normal - non const also) we because at the time of construction we can give some value to const variable.
Second type of Initialization is basically a normal Parametrised constructor. That is used when you are having a object and at the time of creation of object you want to initialize the member variable.

C++, What does the colon after a constructor mean? [duplicate]

This question already has answers here:
Closed 12 years ago.
Possible Duplicates:
Variables After the Colon in a Constructor
C++ constructor syntax question (noob)
I have some C++ code here:
class demo
{
private:
unsigned char len, *dat;
public:
demo(unsigned char le = 5, unsigned char default) : len(le)
{
dat = new char[len];
for (int i = 0; i <= le; i++)
dat[i] = default;
}
void ~demo(void)
{
delete [] *dat;
}
};
class newdemo : public demo
{
private:
int *dat1;
public:
newdemo(void) : demo(0, 0)
{
*dat1 = 0;
return 0;
}
};
My question is, what are the : len(le) and : demo(0, 0) called?
Is it something to do with inheritance?
As others have said, it's an initialisation list. You can use it for two things:
Calling base class constructors
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++ specification, 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.
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++
It's called an initialization list. It initializes members before the body of the constructor executes.
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.
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.
You are calling the constructor of its base class, demo.