Consider this code:
class First
{
public:
int y;
First()
{
y = 90;
}
};
class Second{
public:
int x;
First ob; //if i comment it, value of x is initialised to zero
};
int main()
{
Second obj = Second();
cout << obj.x << endl;
}
This program gives different output when I change it:
If I comment out the line First ob; then value of x is initialised to zero
If the class consists object of class First, then member has garbage value.
If I create object like Second obj; then it has garbage value.
What is the reason it behaves differently when Second class consists only built-in members and when object of another class?
And what is the difference between these statements:
Second obj = Second();
Second obj;
Case 1. When you comment out the line First ob, then Second class becomes a POD type, which can be value-initialized using the syntax Second obj= Second(), and value-initialization, in case of built-in types, means zero-initialized, so x is zero-initialized.
Case 2: When you keep the line First ob, then Second class becomes non-POD type, because First class is non-POD type (as it has user-defined constructor1]). In this case, the syntax Second obj=Second() doesn't initialize built-in data type, if you don't initialize it in the constructor.
Case 3: When write Second obj, then there are again two cases (read them carefully):
obj will be default constructed if there is First ob line. And in this case, x will not be initialized at all, as it is not manually initialized in the constructor of Second.
obj will not be initialized at all if you First ob line is commented out. In this case, Second becomes a POD, and POD types are not initialized if you only write Second obj.
1 See this related topic:
Can't C++ POD type have any constructor?
Have a read through the accepted answer to Do the parentheses after the type name make a difference with new? - without the First object, your class is an 'A', with it your class is a 'B'. Second obj; will default-initialize it, Second obj = Second(); will value-initialize it.
This is one of the areas where C++ has some quite tricky and surprising rules, but I'm not going to cover them in detail here as they've been very well explained in the above question.
What do the following phrases mean in C++: zero-, default- and value-initialization? is also related.
Edit:
The above links only really cover the initialization side of things, not the rules on what makes a class trivial/POD/standard layout/etc. What are Aggregates and PODs and how/why are they special? covers that side of things.
At the risk of being a bit tangential, let me post a modified version of the problem, along with a discussion. Let's consider these classes:
struct Agatha
{
int x;
};
struct Claire
{
Claire() { }
int x;
};
To simplify the initialization syntax and make the example analyzable with Valgrind, let's use dynamic objects:
#include <memory>
#include <iostream>
int main()
{
std::unique_ptr<Agatha> p1(new Agatha); // #1d
std::unique_ptr<Agatha> p2(new Agatha()); // #1v
std::unique_ptr<Claire> q1(new Claire); // #2d
std::unique_ptr<Claire> q2(new Claire()); // #2v
std::cout << p1->x
<< p2->x
<< q1->x
<< q2->x
<< std::endl;
}
Only one of those printing lines is correct! Can you figure out which one?
The answer is #1v. Let's discuss:
Case #1d default-initializes an aggregate, which default-initializes each member, which default-initializes Agatha::x, which leaves it uninitialized.
Case #1v value-initializes a aggregate, which value-initializes all members, and thus value-initializes Agatha::x, and thus zero-initializes it. (This is the good case.)
Case #2d default-initializes a non-aggregate, which calls the default constructor. The default constructor of Claire does not initialize Claire::x, so it is left uninitialized.
Case #2d value-initialization of a non-aggregate also calls the default constructor, and the situation is identical to #2v.
Notice that the whole discussion has nothing to do with PODness, but merely with whether the class is an aggregate, like Agatha, or a non-trivial class type with like Claire. You could add a member of type std::string to Agatha without affecting this example.
Related
So to summarize initialization in C++ , constructor can
initialize primitive types to garbage values in memory
call default constructors on members automatically
initialize member objects with nontrivial constructors in init list
but it cannot init array properly. Ref:
C++: constructor initializer for arrays
My question is: how can I do meaningful initialization in the function body? Before entering the braces, everything has already been initialized in one way or another. The braces only contain cleaning up work, things like cout<<"This object is constructed!"
. Ref:
Use of constructor in C++
What if I want to preprocess parameter before passing it to the constructor of member object? For example:
class Foo{
public:
int x;
Foo(int y){
x = y;
}
};
class Bar{
public:
Foo foo1;
Bar(int y){
int z = superComplicatedOperation(y);
// and all other heavylifting work
foo1 = A(z);
}
};
The code above does not compile because compiler tries to init foo1 before entering the braces in constructor.
I have seen suggestions on solving this problem by providing a default constructor to Foo. So the constructor of Bar will first init foo1 to garbage value, and then assign foo1 to something meaningful in the ctor body. This is also the solution to array initialization in the quoted thread. But isn't "initializing to garbage value" an oxymoron? If the compiler wants me to do that, then it wouldn't complain about having no default ctor.
In a word, ctor is supposed to init the object in a user defined way, but I don't see any way to do a nontrivial init. Am I missing something here?
================================================================================
Clarifications: I am askingI am well aware of the Bar(int y): foo(superComplicatedOperation(y)) syntax. My question is on how to do it nicely. From a design point of view, I believe that codes in superComplicatedOperation() really should belong to the ctors. For example, say Bar takes 2 parameters (int x,int y) to initialize. We can imagine they are xy coordinates on plane. Foo also takes coordinates, but with a different reference frame. Then I need
Bar(int x,int y):foo(f1(a,b),f2(a,b))
And this kind of things quickly get ugly if ctor has a handful of params. Besides, init list looks crowded.
And if ctor body "touches" member objects, it means init list does not set members to a desired state, but rather some default zero-like states(e.g. empty vector). Therefore when calling ctor, first memory for the Bar is allocated, then members objects(subchunks of memory) go through the phases:
garbage value/uninitialized
--- init list or default member ctor --->
default "zero-like" state
--- ctor body -------------------------->
actually desired initial state
So initialization is supposed to be a two step process?
Preprocessing a parameter is fairly trivial:
class Bar
{
Foo foo;
Bar(int y) : foo(superComplicatedOperation(y)) {}
}
or you can use a C++11 delegating constructor:
class Bar
{
Foo foo;
Bar (SuperComplicatedData z) : foo(z) {}
Bar (int y) : Bar(superComplicatedOperation(y)) {}
}
What can you do in a constructor function body? Lots of things: fill containers, configure member objects, set up member object relationships, etc.
[...] constructor can
initialize primitive types to garbage values in memory
call default constructors on members automatically
This is default-initialization, which applies to every member that's not specified in the initializer list and do not have an initializer in the class definition.
initialize member objects with nontrivial constructors in init list
The member initializer list can also initialize primitive members.
but it cannot init array properly. Ref:
C++: constructor initializer for arrays
One of the answers there noted that this can now be done in C++11.
My question is: how can I do meaningful initialization in the function
body? Before entering the braces, everything has already been
initialized in one way or another. The braces only contain cleaning up
work, things like cout<<"This object is constructed!" . Ref:
Use of constructor in C++
Um...no. Frequently there's substantial logic in the constructor body proper. It's definitely not only for "cleaning up". It is true that at the time you enter the constructor body every member is initialized in some way (which may include "not initialized at all" for primitive types), but nothing says you can't assign to them or otherwise modify them. Sometimes you have to if there's some sort of circular dependencies. Sometimes the code to compute the stored value is long enough that it's far more readable inside the constructor body.
The code above does not compile because compiler tries to init foo1
before entering the braces in constructor.
I have seen suggestions on solving this problem by providing a default
constructor to Foo.
Well, you can do : foo1(A(superComplicatedOperation(y)) in your example.
So the constructor of Bar will first init foo1 to
garbage value, and then assign foo1 to something meaningful in the
ctor body. This is also the solution to array initialization in the
quoted thread. But isn't "initializing to garbage value" an oxymoron?
If the compiler wants me to do that, then it wouldn't complain about
having no default ctor.
A default ctor means that a object constructed with it is a valid object. A default ctor doesn't necessarily leave members uninitialized (or "initialized to garbage values"). Consider the default ctor for a std::vector, it had better set its internal state to something that represents an empty vector!
More fundamentally, the job of constructors (default or not) is to establish the object's invariants upon which the functions that operate on the object depend. Some objects may have no invariants, but many do (for instance, vector's internal pointer had better be either a valid pointer or null). If you have no default constructor, the compiler has no way of creating a valid object when no argument is supplied; it can't just give you something "initialized to garbage value", because it has no way of knowing whether that can in fact be a valid object of that type.
You can use an initializer list, as you should for any member variables you need to initialize.
You may prefer to initialize things to sane values and then have a function that can set the values later. It's not as clean, but if you plan to construct many objects and you can precompute the value for many of them at once it may save you some time rather than calculating the values for each object.
#include <iostream>
int superComplicatedOperation(int a)
{
return a * 10;
}
class Foo
{
public:
int x;
Foo(int y)
: x(y)
{
}
};
class Bar
{
public:
Foo foo1;
Bar(int y)
: foo1(superComplicatedOperation(y))
{
}
};
int main()
{
Bar b(7);
std::cout << b.foo1.x << "\n";
return 0;
}
class Bar {
public:
Foo foo1;
Bar(int y) : foo1(superComplicatedOperation(y)) { }
};
You can use the Constructor's initialization list for this work. It passes the values to the Ctors of the members for the first creation, this prevents the creation of a garbage object.
I guess I don't understand. I wrote code in the 1990s which has a constructor with hundreds of lines of code which does all kinds of things: opens files, instantiates dozens of objects which create linked lists and arrays based on processing file data. It certainly can initialize an array properly.
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.
I wrote the following code snippet:
void foo()
{
struct _bar_
{
int a;
} bar;
cout << "Value of a is " << bar.a;
}
and compiled it with g++ 4.2.1 (Mac). The output is "Value of a is 0".
Is it true to say that data members of a struct in c++ are always initialized by default (compared to c)? Or is the observed result just coincidence?
I can imagine that structs in c++ have a default constructor (since a struct and a class is almost the same in c++), which would explain why the data member a of bar is initialized to zero.
The simple answer is yes.
It has a default constructor.
Note: struct and class are identical (apart from the default state of the accesses specifiers).
But whether it initializes the members will depends on how the actual object is declared. In your example no the member is not initialized and a has indeterminate value.
void func()
{
_bar_ a; // Members are NOT initialized.
_bar_ b = _bar_(); // Members are zero-initialized
// From C++14
_bar_ c{}; // New Brace initializer (Members are zero-initialized)
_bar_* aP = new _bar_; // Members are NOT initialized.
_bar_* bP = new _bar_(); // Members are zero-initialized
// From C++14
_bar_ cP = new _bar_{}; // New Brace initializer (Members are zero-initialized)
}
// static storage duration objects
// i.e. objects at the global scope.
_bar_ c; // Members are zero-initialized.
The exact details are explained in the standard at 8.5 Initializers [dcl.init] paragraphs 4-10. But the following is a simplistic summary for this situation.
A structure without a user defined constructor has a compiler generated constructor. But what it does depends on how it is used and it will either default initialize its members (which for POD types is usually nothing) or it may zero initialize its members (which for POD usually means set its members to zero).
PS. Don't use a _ as the first character in a type name. You will bump into problems.
Is it true to say that data members of a struct in c++ are always initialized by default (compared to c)? Or is the observed result just coincidence?
It is a coincidence.
Your code invokes Undefined Behavior; unless you explicitly set the members to 0 they can be anything.
Not an answer, but you might take it to be... if you want to try it:
void foo() {
struct test {
int value;
} x;
std::cout << x.value << std::endl;
x.value = 1000;
}
int main() {
foo();
foo();
}
In your example, the memory already had the 0 value before the variable was created, so you can call it a lucky coincidence (in fact, some OS will zero out all memory before starting a process, which means that 0 is quite a likely value to find in a small short program...), the previous code will call the function twice, and the memory from the first call will be reused in the second one, chances are that the second time around it will print 1000. Note however that the value is still undefined, and that this test might or not show the expected result (i.e. there are many things that the compiler can do and would generate a different result...)
Member variables of a struct are not initialized by default. Just like a class (because a struct is exactly the same thing as a class, only in struct the members are public by default).
Do not rely on this functionality it is non-standard
just add
foo() : a() {}
I can't remember the exact state of gcc 4.2 (i think it is too old) but if you were using C++11 you can do the following
foo()=default;
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.