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;
Related
I want to initialize member object variables in the default constructor of the class.
Let's consider the following,
class ABC {
ABC(int A, int B) {
a = A;
b = B;
}
int a;
int b;
};
class Foo {
Foo();
ABC m_obj1;
};
From the above example, I would like to initialize "obj1" in "Foo::Foo()".
One of the restrictions I have is that I cannot do so in the initializer list, as I need to do some computation before I could initialize the member. So the option available (ASFAIK) is to do so in the body of the default constructor only.
Any inputs, how could I do this?
Edit: Restricting to C++11
Would this be a correct way,
Foo:Foo() {
int x = 10;
int y = 100;
m_Obj1(x, y); //Is this correct? <--------
}
Depending on your exact problem and requirements, multiple solutions might be available:
Option 1: Use a function to do the computations and call Foo constructor
Foo makeFoo()
{
// Computations here that initialize A and B for obj1 constructor
return Foo(A, B)
}
Option 2: Call a function that does the computations and initialize obj1 in Foo member initializer list
ABC initABC() {
// Some computations
return ABC(A, B)
}
Foo() : obj1(initABC()) {}
Option 3: Dynamically allocate obj1, for instance with a std::unique_ptr
Option 4: Use std::optional or an emulated c++11 version as shown by other answers
You simply call the base constructor inside the initializer list of the derived constructor. The initializer list starts with ":" after the parameters of the constructor. See example code!
There is no problem to call functions inside the initializer list itself.
int CallFunc1(int x) { return x*2; }
int CallFunc2(int y) { return y*4; }
class ABC {
public:
ABC(int A, int B):a{CallFunc1(A)},b{CallFunc2(B)} {
std::cout << "Constructor with " << a << " " << b << " called" << std::endl;
}
private:
int a;
int b;
};
class Foo {
public:
Foo(): obj1(1,2){}
Foo( int a, int b): obj1(a, b){}
private:
ABC obj1;
};
int main()
{
Foo foo;
Foo fooo( 9,10);
}
edit:
The best method I can think of for your case is a copy constructor, being more specific on what you need to store helps a lot since if it is just two ints inside a class dynamic allocation is not worth it, the size of the object being constructed makes a difference to what method is best, copy constructors can be slower with much larger data types as the object has to be created twice: once when it is automatically constructed in the parent objects constructor and again when a temporary object is created and all the values have to be copied, which can be slower then dynamically allocating if the size is larger.
As far as I'm aware all objects in a class are automatically initialized/allocated in the constructor so sadly dynamic memory use may be your best bet here.
If you are fine with having the object initialized but empty so you know it is not 'ready' yet you can later fill it with useful data when you would have wanted to initialize it. This can be done with default constructors that set the things inside the object to null values or something similar so you know the object hasn't been properly initialized yet. Then before using the object you can check whether it has been initialized by checking for the null values or by having put a bool in the object that tells you whether it is initialized. Dynamically allocated would still be better in my opinion and makes the code look cleaner overall as all you need to store is a null pointer until the object is needed and then allocated and set to the pointer. It is also very easy to check if the pointer is equal to nullptr to know the state of your object.
Dynamically allocating memory may be a hassle since you have to make sure to get rid of memory leaks and it is slightly slower than using the stack, but it is a necessary skill for c++ since the stack is not enough when making programs that use more than the few available megabytes of data on the stack so if you are doing this simply to avoid the hassle I recommend learning it properly. It would be nice if you could be more specific about what kind of object you want to do this with or if you just want an answer that works for most cases.
eg:
*ABC obj1 = nullptr;
...object is needed
obj1 = new(ABC(constructor stuff));
...obj1 isn't needed
delete obj1;
or c++ automatically deletes it when the program closes.
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.
I have seen people put a parenthesis after the member variable in the initialization list. I wonder why would people do that?
For example, I have a STL container in header file:
class A{
public:
A();
...
private:
vector<string> v;
}
and in source file:
A::A() : v() {}
My question is what is v() and why do people do that since that doesn't look like v is initialized into a value either
That will run the default constructor or initializer (for plain types) for the member. In this context, it will default construct the vector. Since it is the default constructor, it is not necessary here. v would have been default constructed in the absence of an initializer.
class Example {
private:
int defaultInt;
vector<int> defaultVector;
int valuedInt;
vector<int> sizedVector;
public:
Example(int value = 0, size_t vectorLen = 10)
: defaultInt(), defaultVector(), valuedInt(value), sizedVector(vectorLen)
{
//defaultInt is now 0 (since integral types are default-initialized to 0)
//defaultVector is now a std::vector<int>() (a default constructed vector)
//valuedInt is now value since it was initialized to value
//sizedVector is now a vector of 'size' default-intialized ints (so 'size' 0's)
}
};
For kicks and giggles, you could also do thirdVector(vectorLen, value) to get a vector with vectorLen elements with the value value. (So Example(5, 10) would make thirdVector a vector of 10 elements valued 5.)
My question is what is v() and why do people do that since that doesn't look like v is initialized into a value either
This is sometimes done to be more explicit. For non POD types this is not necessary as the default constructor is automatically called for them. If the types default constructor has not been defined or is not accessible, this will cause a compile error.
This makes the most sense for POD types, as their value is undefined when they are not initialized.
struct A
{
int t;
A() : { /* value of t undefined */ }
}
struct A
{
int t;
A() : t() { /* value of t is t's default value of 0 */ }
}
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.
Just a simple question about c++ coding style,
for example, all member variables of a class will be called with the default constructor in the initialization list if we don't do anything else. B default constructor will be called and value will be set to 0, int();
class A
{
A();
private:
B b;
int value;
}
However my question is, even do the default constructor will be called is it a good habit to always do it yourself or does it only add extra lines to the code
A::A() : b(), value() {}
You are touching on one of the sticky corners of C++.
The initialization of POD values in objects is sticky and depends on a few things.
Even I am not sure I can get all the rules correct but I believe #Steve Jessop once wrote an article about here on SO (though I can currently find it).
But some examples:
This class will always be initialized b == false and value = 0.
class A
{
A() : b(), value() {}
B b;
int value;
};
Without an explicit default constructor it is more complex:
Here the compiler will generate a default constructor for you. But how the compiler generated default constructor works depends on the situation. The compiler generated default constructor can do two different forms of initialization and which is used depends on context:
Zero Initialization (All POD members are zero'ed out)
Value Initialization (All POD members are left undefined)
Example:
class B
{
B b;
int value;
};
// Variables of static storage duration (notably globals)
// Will be zero initialized and thus b == false and value = 0
B global; // Initialized
int main()
{
// Object of automatic storage duration or dynamic storage duration
// These will depend on how they are declared.
// Value Initialization (POD re-mains undefined)
B bo1; // b/value undefined
B* bp1 = new B; // b.balue undefined
// Zero Initialization
B bo2 = B(); // b = false, value = 0
B* bp2 = new B(); // b = false, value = 0
// Note: The most obvious syntax for zero initializing a local object
// does not work as it is actually interpreted as a forward
// declaration of a function;
B bo3();
}
It's a good idea to leave the default constructor out. The compiler-generated constructors are much more reliable and maintainable than writing your own versions, and apart from anything else, it's a flat out waste of time to write code the compiler could write for you. If you don't need any construction logic, then don't write a constructor.
However, the int will not be initialized unless you do it, which is a sucky reason to have to write a constructor.
By default int variables are not initialized with a value - you have to do it yourself.
So when you don't set the member variable "value" to some value in the constructor it is left uninitialized.
The default constructor will only be called for class types with default constructors, not for primitives.