What EXACTLY is the difference between INITIALIZATION and ASSIGNMENT ?
PS : If possible please give examples in C and C++ , specifically .
Actually , I was confused by these statements ...
C++ provides another way of initializing member variables that allows us to initialize member variables when they are created rather than afterwards. This is done through use of an initialization list.
Using an initialization list is very similar to doing implicit assignments.
Oh my. Initialization and assignment. Well, that's confusion for sure!
To initialize is to make ready for use. And when we're talking about a variable, that means giving the variable a first, useful value. And one way to do that is by using an assignment.
So it's pretty subtle: assignment is one way to do initialization.
Assignment works well for initializing e.g. an int, but it doesn't work well for initializing e.g. a std::string. Why? Because the std::string object contains at least one pointer to dynamically allocated memory, and
if the object has not yet been initialized, that pointer needs to be set to point at a properly allocated buffer (block of memory to hold the string contents), but
if the object has already been initialized, then an assignment may have to deallocate the old buffer and allocate a new one.
So the std::string object's assignment operator evidently has to behave in two different ways, depending on whether the object has already been initialized or not!
Of course it doesn't behave in two different ways. Instead, for a std::string object the initialization is taken care of by a constructor. You can say that a constructor's job is to take the area of memory that will represent the object, and change the arbitrary bits there to something suitable for the object type, something that represents a valid object state.
That initialization from raw memory should ideally be done once for each object, before any other operations on the object.
And the C++ rules effectively guarantee that. At least as long as you don't use very low level facilities. One might call that the C++ construction guarantee.
So, this means that when you do
std::string s( "one" );
then you're doing simple construction from raw memory, but when you do
std::string s;
s = "two";
then you're first constructing s (with an object state representing an empty string), and then assigning to this already initialized s.
And that, finally, allows me to answer your question. From the point of view of language independent programming the first useful value is presumably the one that's assigned, and so in this view one thinks of the assignment as initialization. Yet, at the C++ technical level initialization has already been done, by a call of std::string's default constructor, so at this level one thinks of the declaration as initialization, and the assignment as just a later change of value.
So, especially the term "initialization" depends on the context!
Simply apply some common sense to sort out what Someone Else probably means.
Cheers & hth.,
In the simplest of terms:
int a = 0; // initialization of a to 0
a = 1; // assignment of a to 1
For built in types its relatively straight forward. For user defined types it can get more complex. Have a look at this article.
For instance:
class A
{
public:
A() : val_(0) // initializer list, initializes val_
{}
A(const int v) : val_(v) // initializes val_
{}
A(const A& rhs) : val_(rhs.val_) // still initialization of val_
{}
private:
int val_;
};
// all initialization:
A a;
A a2(4);
A a3(a2);
a = a3; // assignment
Initialization is creating an instance(of type) with certain value.
int i = 0;
Assignment is to give value to an already created instance(of type).
int i;
i = 0
To Answer your edited Question:
What is the difference between Initializing And Assignment inside constructor? &
What is the advantage?
There is a difference between Initializing a member using initializer list and assigning it an value inside the constructor body.
When you initialize fields via initializer list the constructors will be called once.
If you use the 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.
For an integer data type or POD class members there is no practical overhead.
An Code Example:
class Myclass
{
public:
Myclass (unsigned int param) : param_ (param)
{
}
unsigned int param () const
{
return param_;
}
private:
unsigned int param_;
};
In the above example:
Myclass (unsigned int param) : param_ (param)
This construct is called a Member Initializer List in C++.
It initializes a member param_ to a value param.
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 const member or
Your class doesn't have a default constructor
Initialisation: giving an object an initial value:
int a(0);
int b = 2;
int c = a;
int d(c);
std::vector<int> e;
Assignment: assigning a new value to an object:
a = b;
b = 5;
c = a;
d = 2;
In C the general syntax for initialization is with {}:
struct toto { unsigned a; double c[2] };
struct toto T = { 3, { 4.5, 3.1 } };
struct toto S = { .c = { [1] = 7.0 }, .a = 32 };
The one for S is called "designated initializers" and is only available from C99 onward.
Fields that are omitted are automatically initialized with the
correct 0 for the corresponding type.
this syntax applies even to basic data types like double r = { 1.0
};
There is a catchall initializer that sets all fields to 0, namely { 0 }.
if the variable is of static linkage all expressions of the
initializer must be constant expressions
This {} syntax can not be used directly for assignment, but in C99 you can use compound literals instead like
S = (struct toto){ .c = { [1] = 5.0 } };
So by first creating a temporary object on the RHS and assigning this to your object.
One thing that nobody has yet mentioned is the difference between initialisation and assignment of class fields in the constructor.
Let us consider the class:
class Thing
{
int num;
char c;
public:
Thing();
};
Thing::Thing()
: num(5)
{
c = 'a';
}
What we have here is a constructor that initialises Thing::num to the value of 5, and assigns 'a' to Thing::c. In this case the difference is minor, but as was said before if you were to substitute int and char in this example for some arbitrary classes, we would be talking about the difference between calling a parameterised constructor versus a default constructor followed by operator= function.
Related
Is modify const in constructor C++ standard? I was modifying my struct removing fixed-values(default member initializer) to set it later, at constructor-time but I forget to remove const keyword and noticied it later. To my surprise I didn't get a compile-error, it just worked fine but for test case 2 it give a compiler. How are they different?
test case 1:
struct A
{
const int x = 2;
A()
: x(3)
{
}
};
test case 2:
struct A
{
const int x = 2;
A()
{
x = 3; // compile error! error: read-only variable is not assignable
}
};
In the first example you are initializing the constant variable, in the second you are assigning a value to it, after the variable is default constructed. These are different operations and assignment is not possible for constants after their initialization.
This syntax was added in C++11:
struct A
{
const int x = 2;
In this case the 2 is like a "default" value to be used for initialization of x. If you initialize x in your constructor's initialization list, then that is the value used. However if you did not do so, then 2 is used.
This was added to C++11 to avoid the tedium of having to repeat things in the initialization lists if you had multiple constructors.
The thing is that the object must be fully initialized before the body of your constructor can proceed to start playing with it. This means that there must be reserved enough space in memory to which the instance of new object will be placed.
If you declare data member (int x) to be const, you prohibit yourself from ever changing it's value, once it is created, this means that the value must be set during it's creation.
In example 1 you first create the int x and set it to value 3 (which will reside in memory reserved for your object) and only after that the body of your constructor executes.
In example 2 you create new object (with some value for your int x) and than you try to modify it in your constructor, which is prohibited by the const keyword.
Let us consider the following classes
class test1
{
private:
int a;
int b;
public:
test1():a(0),b(0){}
};
class test2
{
private:
int a;
int b;
public:
test2()
{
a=0;
b=0;
}
};
Now, I know that test1() constructor is the right way to initialize the data members of a class, because in test2() we are performing assignment and not initialization. My questions are:
What might go wrong if we perform assignment instead of initialization?
Doesn't the compiler internally perform assignment in case of test1() constructor? If not, then how are these initialized?
What might go wrong if we perform assignment instead of initialization?
Some class types (and also references and const objects) can't be assigned; some can't be default-initialised; some might be more expensive to default-initialise and reassign than to initialise directly.
Doesn't the compiler internally performs assignment in case of test1() constructor? If no then how are these initialized?
In the case of primitive types like int, there is little or no practical difference between the two. Default-initialisation does nothing, and direct-initialisation and assignment both do essentially the same thing.
In the case of class types, default-initialisation, assignment and direct-initialisation each call different user-defined functions, and some operations may not exist at all; so in general the two examples could have very different behaviour.
For your example there is no real different because you are initializing plain integers.
But assume these integers are objects with constructors, then the compiler would generate the following calls:
// test1
a::copy_constructor(0);
b::copy_constructor(0);
// test2
a::default_constructor();
b::default_constructor();
a::operator = (0);
b::operator = (0);
So depending on your objects test2 could have a huge performance impact. Also by initializing your objects in the initializing lists guaranties that your variables have the data when you enter the constructor. One 'drawback' of the initializer list is that the it is executed in the order that the variables are declared and not in the order of the initializer list, so it could be that you don't want to use the initializer list.
A third variant utilizing direct-list-initialization. Advantages are
variables don't have to be default-initialized in the constructor
the constructor and not the copy/move assignment operator is used to construct the members (as the other answers said, it doesn't matter for the primitive type of this example)
So this is less error-prone than both variants in the question:
#include <iostream>
class Test3
{
public: // Made public so we can easily output the vars. No struct is used to be consistent with question.
int a { 4 }; // No need for separate initialization in constructor.
int b { 2 }; // No need for separate initialization in constructor.
};
int main()
{
const auto t = std::move(Test3()); // Implicitly-declared default constructor + implicitly-declared move assignment operator
std::cout << t.a << t.b;
}
Output is 42.
Constructor-initializers allow initialization of data members at the time of their
creation.
Some programmers prefer to assign initial values in the body of the constructor. However, several data types must be initialized in a constructor-initializer. The following lines summarizes them:
const data members
You cannot legally assign a value to a const variable
after it is created. Any value must be supplied at the
time of creation.
Reference data members
References cannot exist without referring to
something.
Object data members for which there is no default constructor
C++ attempts to initialize member objects using a
default constructor. If no default constructor exists, it
cannot initialize the object.
I want to know why constant data member of a class need to be initialized at the constructor and why not somewhere else? What is the impact of doing so and not doing so?
I also see that only static constant integral data can be initialized inside the class other than that non of the data members can be initialized inside the class.
for eg:- Suppose below is my class declaration
class A{
int a; // This we can initialize at the constructor or we can set this member by calling "vSet" member function
const int b;
static const int c = 10; //This works fine
public:
A();
~A();
void vSet(int a);
int iAdd();
void vDisplay();
};
And the constructor is defined as mentioned below:-
Edited Section: As previous constructor definition example was wrong
A::A():a(1),b(9){}
Please correct me if I am wrong.
Thanks in advance.
A::A(){
a = 1;
b = 9; // Why we need to initialize this only at the constructor.
}
Is not initialization but it is Assignment.
a and b are already constructed and you assign them values in this case. The const qualifier demands that the variable not be changed after its initialization, allowing this assignment would break that contract.
This is Initialization using Member Initialization list.
A::A():a(1),b(9)
{}
You might want to have a look at this answer of mine to know the difference:
What is the difference between Initializing and Assignment inside constructor?
As for another question, regarding only static constant integral data can be initialized inside the class, please read this answer of mine which explains in greater detail:
Why I can't initialize non-const static member or static array in class?
What you're doing is not initialization. It is assignment, so b=9 will NOT even compile, because b is a const, so cannot be assigned any value to it. It should be initialized, and to do that, use member-initialization list as:
A::A() : a(1), b(9)
{ // ^^^^^^^^^^^^ this is called member-initialization list
}
In C++11, you can use in-place initialization as:
class A{
int a = 1; //C++11 only
const int b = 9; //C++11 only
static const int c = 10; //This works fine (both in C++03 and C++11)
//...
};
Because it is constant, it's value cannot be changed. Initializing anywhere else other than the constructor would mean a default initialization, followed by an assignment, which is not permitted for a constant. It would be the equivalent of doing this:
const int i; // OK
i = 42; // Error!
Note that in C++11 it is perfectly OK to do this:
struct Foo {
const int i=42;
const double x = 3.1416;
};
But this follows the same rules, i.e there is no assignment.
A const data is a data that can never be changed. It is initialized once, and then keeps the same value forever.
Because of that, you can't just assign a value to it anywhere.
The constructor, however, is where initialization goes. So here, there is an exception, and you are able to assign a value to your const data. You can do this the classical way, or as many said, as initialization list.
Now, why you can't do this in the class definition (unlike, say, Java) when the variable is not static is another problem, that I know no answer to.
When the body of your constructor is entered, all members and sub-objects are already initialized. The only thing the constructor body can do is to change them – and that is, obviously, not allowed for const members.
You can, however, use an initializer list.
You can't initialize const members outside of a constructor because, well, they're const. By definition, you can't modify them after they've been initialized, and after the object has been constructed, anything that tries to set the value is a modification.
const values are meant to be rvalues, so they cannot appear on the right part of an expression due its constness.
so, when you use this expression on the constructor's body
A::A()
{
a = 1;
b = 9; // Why we need to initialize this only at the constructor.
}
You're using the const value as a lvalue, just as Als mentioned before. The fact is that you're trying to assing a new value to a variable that isn't allowed to change it's value afther it's lifetime had begun.
The correct way to assign values to a constant data member is in the ctor initializer list, that is, BEFORE the lifetime of the member value begins (as mentioned by Nawaz):
A::A() :
a(1),
b(9)
{
}
Finally, on the C++11 standard you're allowed to initialize all data members where it was declared, just like the static const ones.
Of what I know of benefits of using initialization list is that they provide efficiency when initializing class members which are not build-in. For example,
Fred::Fred() : x_(whatever) { }
is preferable to,
Fred::Fred() { x_ = whatever; }
if x is an object of a custom class. Other than that, this style is used even with built-in types for the sake of consistency.
The most common benefit of doing this is improved performance. If the expression whatever is the same type as member variable x_, the result of the whatever expression is constructed directly inside x_ — the compiler does not make a separate copy of the object.
With the other style, the expression whatever causes a separate, temporary object to be created, and this temporary object is passed into the x_ object's assignment operator. Then that temporary object is destructed at the ;. That's inefficient.
Question
Is there any efficiency gain in the following example with using initialization list.
I think there is no gain. The first version calls string's copy constructor and the other calls string's assignment operator (there isn't any temporary thats created). It that correct?
class MyClass
{
public:
MyClass(string n):name(n) { }
private:
string name;
};
class MyClass
{
public:
MyClass(string n)
{
name=n;
}
private:
string name;
};
The second version is calling string's default ctor and then string's copy-assignment operator -- there could definitely be (minor) efficiency losses compared to the first one, which directly calls c's copy-ctor (e.g., depending on string's implementation, there might be useless allocation-then-release of some tiny structure). Why not just always use the right way?-)
I think the only way to initialize const data members is in the initialization list
Eg. in the header:
class C
{
C();
private:
const int x;
int y;
}
And the in the cpp file:
C::C() :
x( 10 ),
y( 10 )
{
x = 20; // fails
y = 20;
}
It's a great way to initialize members that :
are const
don't have a default constructor (it's private)
Remember that there is a distinct difference between a copy constructor and an assignment operator:
the copy ctor constructs a new object using some other instance as a place to get initialization information from.
the assignment operator modifies an already existing object that has already been fully constructed (even if it's only by using a default constructor)
So in your second example, some work has already been done to create name by the time that
name=n;
is reached.
However, it's quite possible (especially in this simple example) that the work done is vanishingly small (probably just zeroing out some data members in the string object) and that the work is optimized away altogether in an optimized build. but it's still considered good form to use initializer lists whenever possible.
We can also perform the constructor delegation via the initialization list.
I am currently taking a c++ course and trying to get a deep understanding of the whole thing.
I came up with some theories, it would be great if somebody could confirm them:
Every variable (local,global,staic,member and non-member) is guaranteed to have its ctor called before first use
The ctors of primitives like int are essentially no-ops, so we have explicitly assign a value, there is no default zero value.
the following classes are semantically the same (and should generate identical code)
class A
{
int n;
};
and
class A
{
int n;
public:
A() : n() {}
};
and
class A
{
int n;
public:
A() { n = int(); }
};
The variable n is in every case still uninitialized.
EDIT:
It seem that I absolutetly underestimated the complexity of this subject, most of my assumptions were wrong. Now Iam still trying to find out the basic rules of object initialisation.
I'm afraid you are wrong. When you say:
int n = int();
Then n (and all other POD types) will zero initialised.
Also, make sure that you are very clear in your mind about the difference between initialisation and assignment - this is quite important in C++:
int n = int(); // initialisation
n = 0; // assignment
You might find this interesting.
The difference between new Foo and new
Foo() is that former will be
uninitialized and the latter will be
default initialized (to zero) when Foo
is a POD type. So, when not using the
form with the parens, the member "a"
can contain garbage, but with the
parens "a" will always be initialized
to 0.
No, the variable is only left uninitialised in the first case.
For a member that is a class with a user-defined constructor, the situation is simple: a constructor is always called.
Built-in types (and 'plain old data' structs) may be left uninitialised, as in your first example. Although they don't have user-supplied constructors, using the construction syntax (your other two examples) initialises them to zero.
The reason for this slightly tricky rule is to avoid undue overhead; for example if you defined:
struct S
{
int array[1024*1024];
};
with the intention of only assigning values as you needed them, you wouldn't want the compiler to whitewash 4Mb of memory with zeroes whenever you construct one.
class A
{
int n;
};
Only memory is allocated, no initialization done for n.
class A
{
int n;
public:
A() : n() {}
};
Here n is initialized with 0.
class A
{
int n;
public:
A() { n = int(); }
};
Here n is first constructed (without any default value),
then int() causes a temp int to be created with value 0
Which is then assigned to n;