Initialization of an object of a class

[ivanov ivan ivanovich]

I would be really appreciated if you can clarify for me the following issue:
Why the ways of initialization in the bellow code work? and they give the same result?

  #include <iostream>
using namespace std;
class ex{
 private:
  float x,y;
 public:
  ex():x(0),y(0){}
  ex(float m,float n):x(m),y(n){};
  ex(float m):x(m),y(0){};
  void show(){
   cout<<"{"<<x<<","<<y<<"}"<<endl;
  }
};
int main(){
 ex x0=(2,3);
 ex x1=3;
 x0.show();
 x1.show();
}

The result is:

{3,0}
{3,0}

Thanks!

[LA48b7Xj]

When you write ex x0=(2,3); (2, 3) evaluates to 3 before calling the constructor so both x0 and x1 are calling the single argument constructor with the value 3.

Maybe you you will get the expected result by removing the = from x0

[ivanov ivan ivanovich]

Thank krako. Can you give me the principle why they work. I mean for initialization we use constructors like ex x0(2,3) or x0(3). I do not understand the reason why assignment works here.

[LA48b7Xj]

class types try to be similar to built in types.

You can do the same thing with int, double, etc.

int x(8);
int y = 9;

[AbstractionAnon]

I do not understand the reason why assignment works here.

The equal sign makes line 15 an assignment statement, not a constructor.
The line 15 parse is:

 
  x1 = <expression>

Where <expression> is (2,3)
The parentheses are effectively ignored.

The comma operator (,) is used to separate two or more expressions that are included where only one expression is expected. When the set of expressions has to be evaluated for a value, only the right-most expression is considered.

http://www.cplusplus.com/doc/tutorial/operators/

[LA48b7Xj]

The equal sign makes line 15 an assignment statement, not a constructor.

The assignment operator isn't invoked here, the constructor is.

[ivanov ivan ivanovich]

Actually, I still don't understand why an assignment can invoke a constructor of a class here.
Or can I understand roughly like that:
For an integer variable, there are two way to initialize its value, they are:

int intvar=2;
int intvar(2);

For an object of a class, the principle is similar. And only one-argument constructor is invoked in this case?

[ivanov ivan ivanovich]

Another issue: why in the following case, the assignment can initialize 3 properties of an object:

#include <iostream>
using namespace std;
class s{
public:
 int x,y,z;
};
int main(){
 s t1={1,2,3};
 cout<<t1.x<<","<<t1.y<<","<<t1.z<<endl;
}

However, if I add constructors for the above class like that:

#include <iostream>
using namespace std;
class s{
public:
 int x,y,z;
 s():x(0),y(0),z(0){};
 s(int x1):x(x1),y(0),z(0){};
};
int main(){
 s t1={1,2,3};
 cout<<t1.x<<","<<t1.y<<","<<t1.z<<endl;
}

, the compiler throws errors like that:

C0_4.cc:10:13: error: in C++98 ‘t1’ must be initialized by constructor, not by ‘{...}’
  s t1={1,2,3};
             ^
C0_4.cc:10:13: error: could not convert ‘{1, 2, 3}’ from ‘<brace-enclosed initializer list>’ to ‘s’

Thanks.