That is not a duplicate. Please read carefully. There are two variables x (of type int and X) and one member is actually declared private, which is used in a constructor. It is about understanding the constructor process in this very specific case.
I am doing a C++ course and I understand the following example that was given. It is about constructors.
#include <iostream>
using namespace std;
class Element {
int value;
public:
Element(int val) {
value = val;
cout << "Element(" << val << ") constructed!" << endl;
}
int Get(void) {
return value;
}
void Put(int val) {
value = val;
}
};
class Collection {
Element el1, el2;
public:
Collection(void) : el2(2), el1(1) {
cout << "Collection constructed!" << endl;
}
int Get(int elno) {
return elno == 1 ? el1.Get() : el2.Get();
}
int Put(int elno, int val) {
if (elno == 1) el1.Put(val);
else el2.Put(val);
}
};
int main(void) {
Collection coll;
return 0;
}
Then they mentioned the following
...
We should also add that there is the following alternation for that
case: when the constructor is divided between the declaration and the
definition, the list of alternative constructors should be associated
with the definition, not the declaration.
It means that the following snippet is correct:
class X {
public:
X(int x) { };
};
class Y {
X x;
public:
Y(int x);
};
Y::Y(int x) : x(1) { };
Can someone explain? Is it really correct? And if yes, how to interpret that? Y has a one-parameter constructor, but no value is passed. x(1) is probably the constructor for the field X x in Y. Is the value of 1 (of x(1)) then passed to Y(int x) automatically although it is declared private in Y?
In the second code snippet, there is no construction actually going on—it's just a definition of the classes and constructors. The only "special" thing about it is that the body of Y's constructor is defined outside the class; it could be in a different file, for example. In this case, it's no different from putting the body directly into the class1:
class Y {
X x;
public:
Y(int x) : x(1) {}
};
When this constructor of Y is invoked, it constructs the member variable x by passing 1 to the X constructor taking int. The constructor of Y doesn't do anything else, i.e. it ignores its own parameter.
The syntax Y::Y used in the original code snippet is standard syntax for defining a member function outside of the class definition. For a non-constructor, it would look like this:
class Foo
{
int bar() const;
};
int Foo::bar() const
{
return 42;
}
1 With the slight difference that when put directly into the class definition, the function is implicitly inline (can be present in more than one translation unit).
Related
In the following code (which is a minimal example based on a much more complex code), a base class defines a struct local to the class. A derived class overrides this definition, but uses also the definition in the base class.
#include <iostream>
struct base {
struct update;
void apply(int& x);
};
struct base::update {
void operator()(int& x)
{
++x;
}
};
void base::apply(int& x) { update{}(x); }
struct deriv : public base {
struct update;
void apply(int& x, int& y);
};
struct deriv::update {
void operator()(int& x, int& y)
{
typename base::update{}.operator()(x);
++y;
}
};
void deriv::apply(int& x, int& y) { update{}(x, y); }
int main()
{
base b;
int x = 1;
b.apply(x);
std::cout << x << std::endl;
int y = 2;
deriv d;
d.apply(x, y);
std::cout << x << ' ' << y << std::endl;
return 0;
}
Compiling with gcc and -Wall -Wextra does not issue any warning.
The output of the code is as expected
2
3 3
However, when analyzing the code with cppcheck, with --enable=all, I get the following:
Checking test.cpp ...
test.cpp:25:24: style: Parameter 'x' can be declared with const [constParameter]
void operator()(int& x, int& y)
^
test.cpp:9:1: error: The one definition rule is violated, different classes/structs have the same name 'update' [ctuOneDefinitionRuleViolation]
struct base::update {
^
test.cpp:24:1: note: The one definition rule is violated, different classes/structs have the same name 'update'
struct deriv::update {
^
test.cpp:9:1: note: The one definition rule is violated, different classes/structs have the same name 'update'
struct base::update {
The first reported error (Parameter 'x' can be declared with const) seems clearly a false positive, as clearly I need non-const reference to be able to modify it, and in fact the code obviously does not compile with a const int&.
But what about the second error on ODR violation? Is this a correct diagnose, and if so why cannot I override the definition of update?
It is also a false positive. There is nothing wrong with declaring a nested class of the same name in the derived class. The two nested classes will be separate entities and each can have one definition per one-definition-rule.
Having the following class with the operator()(string) function and the constructor X(string):
class X
{
public:
string n;
X(string s) : n(s) {}
void operator() (string s)
{
cout << "func";
}
};
Why doesn't X x("a"),y("b"); raise any errors? How does the compiler knows which function to choose?
X x("a"),y("b");
Is equivalent with
X x("a");
X y("b");
So I will focus on
X x("a");
This is a declaration and initialization of the variable x. Only constructors and conversions (*) are considered.
operator() is the function call operator. It is used to call on a object as if it were a function. So it is used on an existing object. E.g.:
X x("asd"); // <-- constructor call
x("yolo"); // <-- operator() call
As a side note I recommend using the brace init syntax {} to initialize objects:
X x{"asd"};
auto x = X{"asd"};
*) As a bonus, here is an example of an user defined conversion operator used:
struct X { };
struct Y
{
operator X() const { return X{}; }
};
auto test()
{
Y y;
X x(y);
// the above line is equivalent with:
X x = static_cast<X>(y.operator X());
}
I used X x(y) to mimic your example, but again, I recommend:
auto test()
{
Y y{};
X x{y};
}
instead
Is there a nice way to have a non static value as default argument in a function? I've seen some older responses to the same question which always end up in explicitly writing out the overload. Is this still necessary in C++17?
What I'd like to do is do something akin to
class C {
const int N; //Initialized in constructor
void foo(int x = this->N){
//do something
}
}
instead of having to write
class C {
const int N; //Initialized in constructor
void foo(){
foo(N);
}
void foo(int x){
//do something
}
}
which makes the purpose of the overload less obvious.
One relatively elegant way (in my opinion) would be to use std::optional to accept the argument, and if no argument was provided, use the default from the object:
class C {
const int N_; // Initialized in constructor
public:
C(int x) :N_(x) {}
void foo(std::optional<int> x = std::nullopt) {
std::cout << x.value_or(N_) << std::endl;
}
};
int main() {
C c(7);
c.foo();
c.foo(0);
}
You can find the full explanation of what works/doesn't work in section 11.3.6 of the standard. Subsection 9 describes member access (excerpt):
A non-static member shall not appear in a default argument unless it
appears as the id-expressionof a class member access expression
(8.5.1.5) or unless it is used to form a pointer to member
(8.5.2.1).[Example:The declaration of X::mem1()in the following example
is ill-formed because no object is supplied for the non-static
memberX::a used as an initializer.
int b;
class X {
int a;
int mem1(int i = a);// error: non-static memberaused as default argument
int mem2(int i = b);// OK; useX::b
static int b;
};
I have template function change that takes a function that takes int and returns an object of type A. So I thought I can use the constructor of A
class A {
int y;
public:
explicit A(int y) : y(2 * y) {
}
};
class B {
A x;
public:
B(int x) : x(x) {
}
template<typename F>
void change(int y, F func) {
x = func(y);
}
};
int main(void) {
B b(7);
b.change(88, A()); // << here
return 0;
}
But the compiler says no matching function for call to ‘A::A()’
How can I make it works?
You can't pass a constructor as a parameter like you are attempting. The C++ standard is very strict on not allowing the memory address of a constructor to be taken.
When you call change(88, A()), you are actually constructing a temp A object (which the compiler should not allow since A does not have a default constructor) and then you are passing that object to the parameter of change(). The compiler is correct to complain, since A does not define an operator()(int) to satisfy the call to func(y) when called in an A object.
To make this work, you need to create a separate function that constructs the A object, and then pass that function to change(), eg:
A createA(int y)
{
return A(y);
}
int main(void) {
B b(7);
b.change(88, createA);
return 0;
}
I know that default constructors initialize objects to their default values, but how do we view these values? If there's a variable of type int, it is supposed to be initialized to 0. But how do we actually view these default values of the constructors? Can anyone please provide a code snippet to demonstrate the same?
Unless specified otherwise, objects are constructed with their default constructor, only if one is available.
And for example ints are not initialized.
This is a common source of huge troubles and bugs, because it can have any value.
So the rule is , always initialise your variables, and for a class you do it in the initialization list
class A
{
private:
int i;
float f;
char * pC;
MyObjectType myObject;
public:
A() : // the initialisation list is after the :
i(0),
f(2.5),
pC(NULL),
myObject("parameter_for_special_constructor")
{}
}
}
In C++, int is not a class and does not have a default (or any other) constructor.
An int is not guaranteed to be initialised to 0.
If you have a class that has an int as an attribute, you should explicitly initialise it in each of the class's constructors (not just the default one).
class sample
{
private:
int x;
public:
sample()
:x(0)
{
}
sample(const int n)
:x(n)
{
}
sample(const sample& other)
:x(other.x)
{
}
// [...]
};
This way you (and users of your class) can "view" the default values.
Good coding practice: write your own constructor, so you know how it will be initialized. This is portable and guaranteed to always have the same behaviour. Your code will be easier to read and the compiler knows how to make that efficient, especially when using the special notation:
class Foo
{
public:
Foo() : i(0), j(0) {}
private:
int i;
int j;
};
AFAIK, if T is a type (not necessarily a class), T() returns a default value for the type T. Here's a small test I ran:
int main()
{
char c = char();
int i = int();
cout << "c = " << hex << (int) c << endl;
cout << "i = " << i << endl;
}
The output is:
c = 0
i = 0
Default constructors do not automatically initialise ints to 0. You can use parentheses to indicate the default value though.
struct X
{
int x;
};
struct X x1; // x1.x is not necessarily 0
struct Y
{
int y;
Y() : y()
{
}
};
struct Y y1; // y1.y will be 0
show your code
And if your int value is a member of class .
you must give it a value in your default constructor func
The default constructor is which can be invoked with 0 parameters. For example
struct X
{
X(int x = 3) //default constructor
{
//...
}
};
It initializes the object to whichever state you code it to initialize. However, if you don't define any constructor at all the compiler will attempt to generate a default constructor for you - which, in turn is equivalent to a constructor with no arguments and no body. That means that all the members of class/struct type will be initialized with their def. ctors and all the members of primitive types like int will remain uninitialized. Please note that I specialy noted that the compiler will attempt to generate a dflt ctor, because it can fail to do so, for example when one or more members of the class do not have default constructors. HTH
AS soon as you have created an object start printing the values such as ob.x.
Thats the only way u can know what default constructor has assigned to the variables.