C++ Constructors Reasoning

Question

Why is the constructor H1 and H2 getting called? Can I get the reason?

class A
{
public:
    int a;
    char b;

    A()
    {
        a = 0;
        b = '\n';
    }

    A(int a)
    {
        cout << "H1";
    }

    A(char c)
    {
        cout << "H2";
    }
};

int main()
{
    int a = 10;
    char c = 'h';
    A ob1 = a;
    ob1 = c;
}

@ScottLawson SSCCEs aren't supposed to do anything http://sscce.org/ they just show a simple example of a concept. The question shows a lack of understanding but at the same time C++ is a complex language with a LOT of tutorials about a LOT of different subjects. A pointer as to what tutorials might help would be more useful. — Philip Couling, Aug 19 '14 at 16:43
@couling agree, basic questions should be answered insightfully — BeyelerStudios, Aug 19 '14 at 16:45

score 4 · Answer 1 · answered Aug 19 '14 at 16:42

A ob1 = a;

This initialises an object of type A from an int value. That uses the constructor A(int), printing H1.

ob1 = c;

This assigns a char value to the object. Since there isn't a suitable assignment operator A::operator=(char), which could assign the value directly, it does this in two stages:

Create a temporary object of type A from the char value
Assign that to ob1 using the implicit copy-assignment operator.

The first stage initialises the temporary using the constructor A(char), printing H2.

Hence, the output is H1H2.

+1. Just to add two subjects to read up on (search on google)... "Implicit Casting" and "Operator Overloading". — Philip Couling, Aug 19 '14 at 16:51

score 0 · Answer 2 · answered Aug 19 '14 at 16:30

class A {
  public:
    int a;
    char b;
    A() {
            a=0;
            b='\n'; 
    }
    A(int a) {
           cout<<"H1";
    }
    A(char c) {
           cout<<"H2";
    } 
}; 

int main() {
    int a=10;
    char c='h';
    A ob1=a; // calls A(int a) because a is typeof int
    ob1=c;  // calls A(char c) because c is typeof char
}

C++ Constructors Reasoning

2 Answers2