Static and Local Scope

Someone says, " I just want to define a int var in a .h file. Who pa Who?

Now we see the following example.

 

Static: Local in YOU

 

 

---------------ah.h-----------------

static int a=1;

------------------------------

 

----------------callseta.cpp--------------

#include"ah.h"

#include<iostream>

using namespace std;

 

void seta()

{

  a=2;

  //cout<<"call seta()"<<endl;

  //cout<<"in call, a="<<a<<endl;

}

--------------------------------------------

 

-----------------main.cpp---------------

#include"ah.h"

#include<iostream>

 

extern void seta();

 

using namespace std;

void main()

{

  cout<<"a="<<a<<endl;

  seta();

  cout<<"after seta()"<<endl;

  cout<<"a="<<a<<endl;

}

-----------------------------------

output  :

a=1
call seta()
in call, a=2
after seta()
a=1
请按任意键继续. . .

 

----------------------------------

this a is local in main.cpp. For include" ", the a is static.

So there are two "int a", one in main.cpp's scope or workspace, the other in callseta.cpp's scope.

They are two different int var, with different address allocated.

They only live in their own world. Any guy outside can not see her beauty.

 

 

Interesting things are:

when you call seta() in main.cpp, the pointer turns to the program space of callseta.cpp.

It is not realized that firstly clone the program to the main.cpp 's program data segament.

The pointer turns onto the orginal and the only one program data segament of seta() in callseta.cpp.

So in that scope, a means his a, not the a in main, which is amazing.

 

posted @ 2016-06-03 20:39  steven_xiu  阅读(127)  评论(0编辑  收藏  举报