I know this is super basic, but will a non pointer value hold its value when it is initialized within a constructor?
class foo {
private:
int bar;
int cool;
public:
foo(int tBar);
~foo();
}
foo::foo(int tBar) {
bar = tBar;
cool = -5;
}
Or in other words, will bar hold tBar's value later on and will cool still be equal to 5 after the construction of the object?
Answer is Yes. I had a misconception that every member variable had to be a pointer!
BUT
When I use more complicated object types (like my own objects) it doesn't work this way. Why is that?
Answer is yes, the object will hold its value
Also I was using crappy objects. That is why they didn't hold value.
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.
I'm relatively new to C++ and I'm wondering if structs are copied in the following case:
struct foo {
int i;
std::vector<int> bar;
}
class Foobar {
foo m_foo;
void store(foo& f) {
this->m_foo = f;
}
}
void main() {
Foobar foobar;
{
foo f;
f.i = 1;
f.bar.insert(2);
foobar.store(f);
}
// will a copy of f still exist in foobar.m_foo, or am I storing a NULL-Pointer at this point?
}
The reason why I am asking this is that I am originally a .NET developer and in .NET structures will be copied if you pass them to a function (and classes are not).
I'm pretty sure it would be copied if store was not declared to take f by reference, but I cannot change this code.
Edit: Updated the code, because I didn't know that the vector.insert would affect my question. In my case I store the struct as a member in a class, not a vector.
So my question really was: will f be copied at this->m_foo = f;?
Short answer: Yes.
Long answer: You'd have to get a pointer to a stack allocated struct and then let that struct go out of scope in order to end up with a dangling reference in your vector... but even then, you wouldn't have stored a NULL. C and C++ pointers are simple things, and will continue to point at a memory location long after that memory location has become invalid, if your code doesn't overwrite them.
It might also be worth noting that std::vector has a decent set of copy and move functions associated with it that will be called implicitly in this case, so the bar vector inside the struct will also be copied along with the simple integer i. Standard library classes tend to be quite well written, but code by other folk has no such guarantee!
Now, as regards your edit:
class Foobar {
foo m_foo;
void store(foo& f) {
this->m_foo = f;
}
}
You will still not have any problems with the foo instance stored in m_foo. This is because this->m_foo = f invokes a copying operation, as m_foo is not a variable of a reference or pointer type. If you had this instead: foo& m_foo then you would run into difficulties because instead of copying a foo instance you are instead copying a reference to a foo instance, and when that instance goes out of scope, the reference is no longer valid.
Yes, the struct will be copied, in the following function:
foos.insert(f);
As a copy is made, you won't be storing a null pointer / null reference.
However, like you've said, it won't be copied when you call store(f); as the function accepts the argument as a reference.
Your edit will still make a copy of Foo. You are assigning one instance of a variable to another instance of a variable. What you aren't doing is assigning one pointer (reference in C#) to another. You could probably do with doing some reading around C++ object instances, pointers, and references.
A copy of f is made during foos.insert(f)
void store(foo& f) {
foos.insert(f);
}
void main() {
{
foo f;
f.i = 1;
f.bar.insert(2);
store(f);
}
// at this place, local variable `f` runs out of scope, it's destroyed and cleaned up
// foos is holding the copy of `f`
}
This is going to sound so basic as to make one think I made zero effort to find the answer myself, but I swear I did search for about 20 minutes and found no answer.
If a private c++ class member variable (non-static) is a pointer, and it is NOT initialized in the constructor (either through an initialization list or an assignment in the constructor), what will its value be when the class is fully instantiated?
Bonus question: If the answer to the above question is anything other than NULL, and I wish to always initialize a particular member pointer variable to NULL, and I have multiple constructors, do I really have to put an explicit initialization for that pointer in every constructor I write? And if so, how do the pros handle this? Surely nobody actually puts redundant initializers for the same member in all their constructors, do they?
EDIT: I wish I could've chosen two answers here. The smart pointers recommended by Bleep Bloop seem to be the elegantest approach, and it got the most up votes. But since I didn't actually use smart pointers in my work (yet), I chose the most illustrative answer that didn't use smart pointers as the answer.
You're thinking correctly. If you don't initialise it, it could be anything.
So the answer to your question is yet, either initialise it with something, or give it a NULL (or nullptr, in the most recent C++ standard).
class A
{
};
class B
{
private:
A* a_;
public:
B() : a_(NULL) { };
B(a* a) : a_(a) { };
};
Our default ctor here makes it NULL (replace with nullptr if needed), the second constructor will initialise it with the value passed (which isn't guaranteed to be good either!).
The value will be uninitialised so yes you do need to explicitly initialise it to nullptr.
Using smart pointers (std::unique_ptr, std::shared_ptr, boost::shared_ptr, etc.) would mean that you don't need to do this explicitly.
the value of any uninitialized pointer is always garbage, it's some random memory address.
in your constructors, you can use initializer lists to initialize your pointer
simply
MyClass::MyClass() : myPointer(nullptr)
{
}
trying to reference an uninitialized pointer triggers undefined behavior.
so ALWAYS initialize your pointer.
Value will be undefined.
You may have one "ultimate" ctor which will initialize all fields and add "short-cut" ctors with only part of parameters, which will pass these params to ultimate ctor along with default values for the rest of params.
Even if the most voted answer is technically correct, I would suggest better approach is to initialise variable in class itself.
class A
{
};
class B
{
private:
A* a_ = nullptr;
public:
B() : { };
B(a* a) : a_(a) { };
};
In C++, you are allowed to initialize values to member variables using the initialization list syntax. See this:
class AnyClass
{
};
class Xyz
{
int n;
AnyClass *p;
Xyz() : n(55), p(new AnyClass())
{
// constructor body
}
~Xyz() { delete p; }
};
The value of n becomes 55 and the pointer to the newly created AnyClass is initialized into p. Note that they are not assigned to n and p. They are initialized into the member variables.
When using new, you should delete it using delete When using new[], you should use delete[].
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.
I have edited this from my real code, so that it is a little easier to understand.
The base class:
class MWTypes
{
public:
virtual long get() { return (0); }
};
The derived class: (There are going to be other classes like char, double etc etc . . .)
class TypeLong : public MWTypes
{
public:
TypeLong(long& ref) : m_long(ref) {}
~TypeLong();
long get() { return m_long; }
private:
long& m_long;
};
and the storage class:
class RowSet
{
public:
void addElememnt(MWTypes elem);
MWTypes getElement();
std::vector<MWTypes> getVector() { return m_row; }
private:
std::vector<MWTypes> m_row;
};
How it is called:
for (i = 0; i < NumCols; i++) // NumCols is 3 on this instance
{
switch(CTypeArray[i]) // this is an int which identifies the type
{
case SQL_INTEGER:
{
long _long = 0;
TypeLong longObj(_long);
MWTypes *ptr = &longObj;
// some SQL code goes here that changes the value of _long,
// there is no need to include it, so this will do.
_long++;
// I now want to save the data in a vector to be returned to the user.
rowSet.addElememnt(*ptr);
///////////////////////////////////////////////
// some code happens here that is irrelevant //
///////////////////////////////////////////////
// I now want to return the typr I have saved in the vector,
// I THINK I am doing this right?
MWTypes returned = rowSet.getElement();
// lastly I want to get the value in the returned type
long foo = returned.get();
///////////////////////////////////////////////
// some code happens here that is irrelevant //
///////////////////////////////////////////////
I think I am on the right lines here. The value of 'foo' is always 0. I have a feeling this could be the way Im storing in the vector, or it could be the base virtual function, as it always returns 0.
If I remove the return in my base class I get LNK2001 errors.
MWTypes returned = rowSet.getElement();
// lastly I want to get the value in the returned type
long foo = returned.get();
should be
MWTypes* returned = &rowSet.getElement();
// lastly I want to get the value in the returned type
long foo = returned->get();
or
MWTypes& returned = rowSet.getElement(); // actually illegal, but MSVC will let you do
// lastly I want to get the value in the returned type
long foo = returned.get();
Indeed, polymorphic calls must be made via a pointer or a reference.
EDIT: this is not your only problem. The fact that the vector stores objects (and not pointers) will slice the objects and destroy their type information.
See this faq entry for additional info to help you solve your problem and understand how virtual functions are called.
The fundamental problem is that you are making copies of your objects of type MWTypes, thus losing their particular subclass. If you want to use an object of an unknown subclass of the base class, then you can only use a pointer or reference to the base type, not an actual instance of it.
Not providing an implementation of the function "get" as ascanio's code shows (making the function "pure virtual") would prevent you from being able to make this copying mistake, because the compiler would not let you instantiate the class MWTypes if you did that (it would say the class is "abstract").
You are suffering from slicing since your collection stores copies of the base type. Whenever you store something into the vector, your code just slices off the base part and it forgets its original type.
To fix this, you could store pointers to the base: std::vector<MWTypes*>, but then you have to manage your instances correctly to avoid memory leaks.
class RowSet
{
public:
// addElement assumes responsibility for the memory allocated for each 'elem'
void addElement(MWTypes* elem);
MWTypes* getElement();
std::vector<MWTypes*> getVector() { return m_row; }
// Destructor calls delete on every pointer in m_row
~RowSet();
private:
std::vector<MWTypes*> m_row;
};
Then you need to fix your code which calls addElement() to create new instances, and to get the long back again:
rowSet.getElement()->get();
You're problem lies with this function void addElememnt(MWTypes elem);. It should be either void addElememnt(MWTypes* elem); or void addElememnt(MWTypes& elem);. This is because by having an argument to be passed by-value, it loses it's polymorphism. The passing by-value calls the copy constructor of the base class and ONLY copies the contents of the base class (and the vtable) ignoring the rest from the derived class.
Also, if you need to store values of a certain base-class type, you need to consider using a list of pointers of the base-class type.
The problem lies here:
class RowSet
{
public:
void addElememnt(MWTypes elem);
You are taking elem by value, not by pointer or by reference, so the TypeLong subobject is sliced away, here: (reference: What Is The Slicing Problem in C++?)
TypeLong longObj(_long);
MWTypes *ptr = &longObj;
_long++;
rowSet.addElememnt(*ptr);
You need to change addElement to take a reference or a pointer.
Your vector, getElement, and addElememnt parts all invoke object slicing since they store the base object by value. You need to work with pointers or references in order to use runtime polymorphism.
In this case either a boost::ptr_vector or a vector of shared_ptr is probably what you want.