Which constructor will be called? [duplicate] - c++

This question already has answers here:
Default constructor with empty brackets
(9 answers)
What does X f() mean?
(4 answers)
Closed 4 years ago.
Why there is no output?
Why doesn’t an ambiguity error occur?
#include <iostream>
using namespace std;
class A{
public:
A(){
cout << "Class A";
}
A(int a = 0){
cout << "A";
}
};
int main() {
A a();
return 0;
}
So when we create an object “a” of type “A” A a();, compiler calls a constructor without parameters, doesn’t he?
In C++, We can have more than one constructor in a class with same name, as long as each has a different list of arguments. A constructor is called depending upon the number and type of arguments passed.
While creating the object, arguments must be passed to let compiler know, which constructor needs to be called.

Related

classname classobject(); does not invoke any constructor [duplicate]

This question already has answers here:
My attempt at value initialization is interpreted as a function declaration, and why doesn't A a(()); solve it?
(5 answers)
Closed 3 years ago.
If I have a following situation:
#include <iostream>
using namespace std;
class A {
public:
A() {
cout << "Inside A" << endl;
}
};
int main() {
A a();
return 0;
}
Why is the constructor not invoked?
If something looks like a function declaration, the C++ standard requires it be treated as a function declaration.
A a(); does not default-construct an object a of type A. It declares a function a that takes no input parameters and returns an A object as output.
To default-construct a variable a, you need to drop the parenthesis:
A a;
Or, in C++11 and later, you can use curly braces instead of parenthesis:
A a{};

constructor resolution, default constructors and () [duplicate]

This question already has answers here:
My attempt at value initialization is interpreted as a function declaration, and why doesn't A a(()); solve it?
(5 answers)
Closed 5 years ago.
#include <iostream>
using namespace std;
class A{
public:
A(){
cout << "Class A!";}
};
int main()
{
A a();
}
the above code does not call the constructor A :: A() even though they have the same input parameters (none). however if, in the main function, I remove the parenthesis from
A a();
it calls the constructor.
so what is the difference between A a; and A a();
i believe the question here is very similar maybe even the same however if some could explain in simpler terms I would be very grateful.
Do the parentheses after the type name make a difference with new?
would declaring A a(); ever call a constructor, under any circumstances?
do parameterless constructors exist in c++, or is that the same as a default constructor?
With
A a();
you declare a as a function taking no arguments and returning an A object.

c++ nested constructors call issue [duplicate]

This question already has answers here:
My attempt at value initialization is interpreted as a function declaration, and why doesn't A a(()); solve it?
(5 answers)
Closed 5 years ago.
I have 2 test classes:
class B {
public:
B(int i) {
qDebug() << "B constructor ";
}
};
class A {
public:
A(B b) {
qDebug() << "A constructor ";
}
};
Now i want to create A object with B, thats what i do:
int i = 5;
A test (B(i)); //does not work
Code compiles without errors, but second line doesnt execute at all. I've made some tests, and code below works well:
int i = 5;
A test (B((int)i)); //works
A test (B(5)); //works
So, i guess compiler cant interpret 'i' as int inside B constructor call, but why?
This:
A test(B(i));
is the same as the:
A test(B i);
which is a function declaration, not a call to constructor due to a most vexing parse.
There is a rule that states (S. Meyers, "Effective Modern C++"):
anything that can be parsed as a declaration must be interpreted as
one
To avoid that use braced initialization (instead of parentheses ()) as functions can't be declared with {} braces:
A test{B(i)};
That being said there are no "nested constructors calls" in your example.

Why Constructor is not Calling with Function call Operators [duplicate]

This question already has answers here:
Default constructor with empty brackets
(9 answers)
Closed 7 years ago.
Sorry I was unable to find any keyword to search for this problem. My code is below
#include <iostream>
using namespace std;
class Name {
public:
Name() {
cout << "Asif";
}
};
int main() {
Name obj(); // why Constructor not calling here?
return 0;
}
If it's not calling the constructor then what process is running in this code?
Name obj(); // why Constructor not calling here?
This is a function declaration, returning a Name object
Just do
Name obj;

What's the difference in instantiating variable with vs. without braces in C++? [duplicate]

This question already has answers here:
When to use the brace-enclosed initializer?
(3 answers)
Instantiate class with or without parentheses? [duplicate]
(1 answer)
Closed 9 years ago.
#include <iostream>
using namespace std;
struct CTest
{
CTest() { cout << "Constructor called"; }
CTest(string s) { cout << "Any constructor with parameters"; }
};
int main () {
CTest t1;
CTest t2{};
}
I come from the Java world and there t1 would just have been declared which definitely isn't the case here since both both lines call the constructor of CTtest. In this case, t1 calls the overwritten default constructor as well as t2. Are there any cases where it actually makes a difference or can we always omit the braces?
Maybe it's just me, but I couldn't find any hint on that. There are only discussions about when to use braces vs. parentheses (vs. value vs. copy constructor).
When the only constructor for a class is its default constructor then doing initialization with curly braces doesn't matter:
CTest t1;
CTest t2{};
Are the same.
Its only once you have other constructors that take parameters that putting values for those parameters inside of {} that you are doing something new.