some question about static member

[Qin Chen]

Hi, all!
I need some experieced coder to explain what the following code will
produce, thank you!

#inlcude <iostream>
using namespace std;

class WithStatic{
static int x;
static int y;
public:
void print(void);
};
void WithStatic:rint(void)
{
cout << "x = " << x << endl;
cout << "y = " << y << endl;
}
int x = 100;
int WithStatic::y = x + 1;
int WithStatic::x = 10;

int main()
{
WithStatic ws;
ws.print();
}

 

[jeffc]

Hi, all!
I need some experieced coder to explain what the following code will
produce, thank you!

It produces a compiler error. Either you didn't try it yourself, or you
didn't cut and paste your actual code here. Is this a homework assignment?
Why do you need an instance of WithStatic?

 

[tom_usenet]

Hi, all!
I need some experieced coder to explain what the following code will
produce, thank you!

#inlcude <iostream>

using namespace std;

class WithStatic{
static int x;
static int y;
public:
void print(void);
};
void WithStatic:rint(void)
{
cout << "x = " << x << endl;
cout << "y = " << y << endl;
}
int x = 100;
int WithStatic::y = x + 1;
int WithStatic::x = 10;

int main()
{
WithStatic ws;
ws.print();
}

That code must produce:

x = 10
y = 11

This is because the initializer for a static variable is in the scope
of the class. Members of the class can be used without qualification,
and hide names from enclosing scopes to the initialization, so the "x
+ 1" refers to "WithStatic::x + 1".

Because WithStatic::x is statically initialized but WithStatic::y is
dynamically initialized (with the non-constant expression "x + 1"),
the fact that it appears before the x initialization doesn't matter -
it will still get the correct value of WithStatic::x. This is similar
to:

extern int i;
int j = i + 1; //dynamic initialization happens later
int i = 10; //static initialization happens first

Here j == 11.

Tom