struct in C++ ......

To C++ Expert,

  I got a code as follows :
------------------
 struct A{
  A(A* a_ptr){ cout << "woof" << endl;}
 } ;

 A* A(A* a_ptr){ cout <<"meow" << endl ;}

 struct A* a1=A(0) ;  //line 1
 struct A* a2 = new struct A(0) ; //line 2
 struct A a3 = A(0) ; //lin3
------------------
 The code does compile and work. But I don't understand :
 
 1. How come the function "A* A(A* a_ptr)" can
     take "0" as  an argument ? i.e. How come line 1
     works ?
 2. How come it is "new struct A(0)" not "new A(0)" ?
     Also, how come the constructor "A(A* a_ptr)"
     can take "0" as an argument ? i.e. How come
     line 2 works ?

    Thanks a lot !!!  
 
LVL 1
meow00Asked:
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jcwlcCommented:
1. A takes a pointer to struct A, the pointer can be null, in c++ 0=NULL
 good practice is to check for null pointers... or at least assert in debug mode. eg: ASSERT(a_ptr)
2. In c++ new allocates memory according to the specified type, not object. A is not a type, struct is... If You typedef struct A... myStructType, then you could write myStructType A = new myStructType.

regards,

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yasser_helmyCommented:
In C++ (as in C), 0 represents NULL.. the type conversion between int and A* is done automatically..

I think in c++, you can use both syntax; you can use "new struct A(0)" and/or"new A(0)".. try it and see if it works.. :)
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meow00Author Commented:
Hi,

   I tried "struct foo f4 = new f00(0)" and "foo f4 = new f00(0)",
 neither of them compiles. I don't understand that why it works for class but not for struct ....?? Thanks.

meow ......

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yasser_helmyCommented:
you need:
struct foo* f4= new foo(0);
new returns a pointer to the structure not the structure itself.
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yasser_helmyCommented:
accept an answer please
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