local variable

Question

as i know, after fun(&a); 

the local variable will be destroyed because function is over and pop out from stack then

how *a can access j variable which is already deleted?

Answer 1

We are not accessing variable j after exiting from function.

Parameter is passed as reference.

Address of a is passed to the function.

Inside the function,value nof j is copied to the address of a.

After exiting drom function, local variable j is deleted, but its value is already stored in address of a.

Thats how we accessed the value

Comments

I didn't get you  ,please explain with pictures. And stepwise

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See program is simply flow in normal way .The only logic here is

    *p=&j;

p contains address of a. So * of p will replacing value of a by address of j.So variable a contains address of j.

 printf("%d\n",**p);

will print value at address j.So print 4.And same for

 printf("%d\n",*a);

will print the same value i.e. 4.

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yes it will be 4,4

As it is call by reference. rt?

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@manoj sir, after printing first 4 by statement **p. After this fun function is over and poped out from stack.

Now my doubt is variable remain into existence. or deallocate the memory which is assigned to local variable j and p.

It is big doubt for me please help!!!!!!

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even it is call by reference but once fun will be over it is popped out so j will be popped then how 4 will be printed if j is not there but pointer will still point to that de-allocated portion as we have not used free ....??

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Is the content remains, at the address

after function pop out ?

Only variable name is remove from address.

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In C  memory is reused once the function is popped out.

http://stackoverflow.com/questions/16980406/when-is-memory-deallocated-in-c