Initializer List is used in initializing the data members of a class. The list of members to be initialized is indicated with constructor as a comma-separated list followed by a colon. Following is an example that uses the initializer list to initialize x and y of Point class.
Example
C++
#include <iostream>
using namespace std;
class Point {
private:
int x;
int y;
public:
Point(int i = 0, int j = 0): x(i), y(j) {}
int getX() const { return x; }
int getY() const { return y; }
};
int main()
{
Point t1(10, 15);
cout << "x = " << t1.getX() << ", ";
cout << "y = " << t1.getY();
return 0;
}
|
The above code is just an example for syntax of the Initializer list. In the above code, x and y can also be easily initialed inside the constructor. But there are situations where initialization of data members inside constructor doesn’t work and Initializer List must be used. The following are such cases:
1. For Initialization of Non-Static const Data Members
const data members must be initialized using Initializer List. In the following example, “t” is a const data member of Test class and is initialized using Initializer List. Reason for initializing the const data member in the initializer list is because no memory is allocated separately for const data member, it is folded in the symbol table due to which we need to initialize it in the initializer list.
Also, it is a Parameterized constructor and we don’t need to call the assignment operator which means we are avoiding one extra operation.
Example
C++
#include<iostream>
using namespace std;
class Test {
const int t;
public:
Test(int t):t(t) {}
int getT() { return t; }
};
int main() {
Test t1(10);
cout<<t1.getT();
return 0;
}
|
2. For Initialization of Reference Members
Reference members must be initialized using the Initializer List. In the following example, “t” is a reference member of the Test class and is initialized using the Initializer List.
Example
C++
#include<iostream>
using namespace std;
class Test {
int &t;
public:
Test(int &t):t(t) {}
int getT() { return t; }
};
int main() {
int x = 20;
Test t1(x);
cout<<t1.getT()<<endl;
x = 30;
cout<<t1.getT()<<endl;
return 0;
}
|
3. For Initialization of Member Objects that do not have a Default Constructor
In the following example, an object “a” of class “A” is a data member of class “B”, and “A” doesn’t have a default constructor. Initializer List must be used to initialize “a”.
Example
C++
#include <iostream>
using namespace std;
class A {
int i;
public:
A(int);
};
A::A(int arg)
{
i = arg;
cout << "A's Constructor called: Value of i: " << i
<< endl;
}
class B {
A a;
public:
B(int);
};
B::B(int x) : a(x)
{
cout << "B's Constructor called";
}
int main()
{
B obj(10);
return 0;
}
|
Output
A's Constructor called: Value of i: 10
B's Constructor called
If class A had both default and parameterized constructors, then Initializer List is not a must if we want to initialize “a” using the default constructor, but it is must to initialize “a” using the parameterized constructor.
4. For Initialization of Base Class Members
Like point 3, the parameterized constructor of the base class can only be called using the Initializer List.
Example
C++
#include <iostream>
using namespace std;
class A {
int i;
public:
A(int );
};
A::A(int arg) {
i = arg;
cout << "A's Constructor called: Value of i: " << i << endl;
}
class B: A {
public:
B(int );
};
B::B(int x):A(x) {
cout << "B's Constructor called";
}
int main() {
B obj(10);
return 0;
}
|
Output
A's Constructor called: Value of i: 10
B's Constructor called
5. When the Constructor’s Parameter Name is the Same as Data Member
If the constructor’s parameter name is the same as the data member name then the data member must be initialized either using this pointer or Initializer List. In the following example, both the member name and parameter name for A() is “i”.
Example
C++
#include <iostream>
using namespace std;
class A {
int i;
public:
A(int);
int getI() const { return i; }
};
A::A(int i) : i(i)
{
}
int main()
{
A a(10);
cout << a.getI();
return 0;
}
|
6. For Performance Reasons
It is better to initialize all class variables in the Initializer List instead of assigning values inside the body. Consider the following example:
Example
C++
class MyClass {
Type variable;
public:
MyClass(Type a) {
variable = a;
}
};
|
Here compiler follows following steps to create an object of type MyClass
1. Type’s constructor is called first for “a”.
2. Default construct “variable”
3. The assignment operator of “Type” is called inside body of MyClass() constructor to assign
variable = a;
4. And then finally destructor of “Type” is called for “a” since it goes out of scope.
Now consider the same code with MyClass() constructor with Initializer List
C++
class MyClass {
Type variable;
public:
MyClass(Type a):variable(a) {
}
};
|
With the Initializer List, the following steps are followed by compiler:
1. Type’s constructor is called first for “a”.
2. Parameterized constructor of “Type” class is called to initialize: variable(a). The arguments in the initializer list are used to copy construct “variable” directly.
3. The destructor of “Type” is called for “a” since it goes out of scope.
As we can see from this example if we use assignment inside constructor body there are three function calls: constructor + destructor + one addition assignment operator call. And if we use Initializer List there are only two function calls: copy constructor + destructor call. See this post for a running example on this point.
This assignment penalty will be much more in “real” applications where there will be many such variables. Thanks to ptr for adding this point.
Parameter vs Uniform Initialization in C++
It is better to use an initialization list with uniform initialization {} rather than parameter initialization () to avoid the issue of narrowing conversions and unexpected behavior. It provides stricter type-checking during initialization and prevents potential narrowing conversions
Code using parameter initialization ()
C++
#include <iostream>
class Base {
char x;
public:
Base(char a)
: x{ a }
{
}
void print() { std::cout << static_cast<int>(x); }
};
int main()
{
Base b{ 300 };
b.print();
return 0;
}
|
In the above code, the value 300 is out of the valid range for char, which may lead to undefined behavior and potentially incorrect results. The compiler might generate a warning or error for this situation, depending on the compilation settings.
Code using uniform initialization {}
By using uniform initialization with {} and initializing x with the provided value a, the compiler will perform stricter type-checking and issue a warning or error during compilation, indicating the narrowing conversion from int to char.
Here is code with uniform initialization {} , which results in a warning and thus better to use
C++
#include <iostream>
class Base {
char x;
public:
Base(char a)
: x{ a }
{
}
void print() { std::cout << static_cast<int>(x); }
};
int main()
{
Base b{ 300 };
b.print();
return 0;
}
|
main.cpp: In function ‘int main()’:
main.cpp:17:17: error: narrowing conversion of ‘300’ from ‘int’ to ‘char’ [-Wnarrowing]
17 | Base b{ 300 }; // Using uniform initialization with {}
| ^
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
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Last Updated :
15 Nov, 2023
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