//the following two code, compile and run the error, will show that the Process is terminated due to StackOverflowException.
//what reason be excuse me? I am familiar with such as stack memory is not enough, please master guiding

Class Program
{
The static void Main (string [] args)
{
A. A=new A ();
Console. WriteLine (" Ok ");
}
}
Class A: the Program
{
B B=new B ();
}
Class B: A {}

&

Class Program
{

B B=new B ();
The static void Main (string [] args)
{
A. A=new A ();
Console. WriteLine (" Ok ");
}
}
Class A: the Program {}
Class B: A {}

CodePudding user response:

Class A
{
A, b=new A ().
}
This will trigger a stack overflow,

Because each to create A new instance, the instance initialization class member b, must be another A new instance,
Thus layers of creating endlessly, until the stack overflow,

CodePudding user response:

For the first time see this application,

Is actually in the circulation, although is different instances, but has been created, you will not survive,

You can debug the see

CodePudding user response:

Because subclass initialization, must run in to the parent class constructor side first,

CodePudding user response:

subclass initialized first perform the superclass constructor,
In fact you code 1 and code 2, is the same,
There is no difference between nature,

Only one on new parent class A, B
One on top of another top parent Program in new B
New subclasses so as long as the parent class, but a subclass has inherited from the superclass,
Got caught up in the subclass initialization time, first initialize the parent class, and then the parent class new subclasses, subclass continue to initialize the infinite loop,

CodePudding user response:

refer to 4th floor are nu month god reply:

subclass initialized first perform the superclass constructor,
In fact you code 1 and code 2, is the same,
There is no difference between nature,

Only one on new parent class A, B
One on top of another top parent Program in new B
New subclasses so as long as the parent class, but a subclass has inherited from the superclass,
Got caught up in the subclass initialization time, first initialize the parent class, and then the parent class new subclasses, subclass continue to initialize the infinite loop,

That is what I put two new objects are inside the main method is no problem

CodePudding user response:

The

reference 3 floor XBodhi. Response:

because subclass initialization, you must first run in to the parent class constructor,

Why after A don't inherit the Program can be up again?

CodePudding user response:

The

reference 5 floor [* Lcw

reply:]

Quote: refer to 4th floor are nu month god reply:

subclass initialized first perform the superclass constructor,
In fact you code 1 and code 2, is the same,
There is no difference between nature,

Only one on new parent class A, B
One on top of another top parent Program in new B
New subclasses so as long as the parent class, but a subclass has inherited from the superclass,
Got caught up in the subclass initialization time, first initialize the parent class, and then the parent class new subclasses, subclass continue to initialize the infinite loop,

That is what I put two new objects are the main methods in no problem

The code

CodePudding user response:

references on 7th floor is wrath of god reply:

Quote: refer to fifth floor [* Lcw

reply:]

Quote: refer to 4th floor are nu month god reply:

subclass initialized first perform the superclass constructor,
In fact you code 1 and code 2, is the same,
There is no difference between nature,

Only one on new parent class A, B
One on top of another top parent Program in new B
New subclasses so as long as the parent class, but a subclass has inherited from the superclass,
Got caught up in the subclass initialization time, first initialize the parent class, and then the parent class new subclasses, subclass continue to initialize the infinite loop,

That is what I put two new objects are the main methods in no problem

The code to the

Class Program
{
The static void Main (string [] args)
{
A. A=new A ();
B B=new B ();
Console. WriteLine (" Ok ");
}
}
Class A: the Program
{
//B B=new B ();
}
Class B: A {}

CodePudding user response:

refer to the eighth floor [* Lcw

reply:]
Class Program
{
The static void Main (string [] args)
{
A. A=new A ();
B B=new B ();
Console. WriteLine (" Ok ");
}
}
Class A: the Program
{
//B B=new B ();
}
Class B: A {}

Did you not comment, //B B=new B ();

CodePudding user response:

If need to explain here,
You go in the Program statement B B=new B ();
So because of B: A and B: Program
So, every time the Program initialization B B=new B ();
References will be nested stackoverflow,

But why B B=new B (); In the Main not,
This is because the Program of the other methods are static, not initialized again,
You can breakpoint debugging knew,
And you can directly in the Program to declare static B=new B ();
That would not produce a nested loop, for the second time will not be instantiated at all,

CodePudding user response:

references to the tenth floor is nu month god reply:

need to explain here, if
You go in the Program statement B B=new B ();
So because of B: A and B: Program
So, every time the Program initialization B B=new B ();
References will be nested stackoverflow,

But why B B=new B (); In the Main not,
This is because the Program of the other methods are static, not initialized again,
You can breakpoint debugging knew,
And you can directly in the Program to declare static B=new B ();
That would not produce a nested loop, for the second time will not be instantiated, at

references to the tenth floor is nu month god reply:

need to explain here, if
You go in the Program statement B B=new B ();
So because of B: A and B: Program
So, every time the Program initialization B B=new B ();
References will be nested stackoverflow,

But why B B=new B (); In the Main not,
This is because the Program of the other methods are static, not initialized again,
You can breakpoint debugging knew,
And you can directly in the Program to declare

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