pointers trivia

[Encrypted]

i have two questions right now in my mind :

question - 1 :

please have a look at this code :

void function (int recd)
{
   printf ("%d", recd);
}
void main()
{
  int p=10;
  function(p);
}

here...when p is passed to function, the thing that actually takes place is :
a new variable recd is created and value of p is copied to recd (i mean a copy of p is created...consu ming another 2 bytes)...

now please check the following code :

void function (int *recd)
{
   printf ("%d", *recd);
}
void main()
{
  int *p;
  *p=10;
  function(p);
}

does the same thing happen here?? i mean here too.. new pointer variable recd is created and value (an address) stored in pointer p is copied to recd???

void function (int *recd)
{
  printf ("%d", *recd);
}

void main()
{
   int p=10;
   function(&p);
}

and here too, does the same thing happens???
please throw some light on it... :)


question - 2:

void main()
{
   int **p;
   **p = 10;   
}

in the above code, where the hell is 10 stored???
what exactly happen in above code??
my guess is : its stored in *p...
i mean **p has a garbage which is taken as *p...and *p has a garbage which is taken as the address of 10....is this right or a garbage?? :P :)

[weaknessforcats]

Answer to Question 1: All variables passed to functions are copied. This is called pass by value and it is all there is in C. The same is true of C++ except that there is a reference function argument that is not a copy of the variable used on the call but becomes a second name for that variable. The reference was developed to avoid the error-prone C pointer syntax.

The fact that the variable used on the call is a pointer is irrelevant. The function has a copy of the pointer. It's just that by using the contents of the pointer a variable outside the function can be accessed but that is a feature of the pointer's data and not the pointer itself. In A and C++ a pointer is a variable like any other.

Answer to Question 2: Indeterminate behavior. It depends on your compiler. Visual Studio.NET generates a warning that an uninitialized variable is being used, but it's just a warning. Ultimately, you are responsible for memory.

[Encrypted]

heyyy...thnxx..
i got your points ... :)

[Ganon11]

In your second section of code, I believe you will suffer the same errors as the final code portion: that is, you have:

int *p;
*p = 10;

p has no value (since you have not used new (C++) or malloc (C)), and so you are putting 10 into some random location in memory.