CODEWITHSUNDEEP
c++c
user2559696
user2559696

Reputation: 109

address of a pointer variable

I'm currently learning c language, and I bumped into this code. ptr is already a pointer type of variable, so what is the effect of the & operator on it, cause I know that usually the operator uses to get the address of non-pointer variable.

struct name {
  int a; float b; char c[30];
};
int main()
{
  struct name *ptr;
  int i,n;
  printf("Enter n: ");
  scanf("%d",&n);
  ptr = (struct name*)malloc(n*sizeof(struct name));
  /* Above statement allocates the memory for n structures with pointer ptr pointing to       base address */ 
  for(i=0; i<n; ++i) { 
    printf("Enter string, integer and floating number respectively:\n");
    scanf("%s%d%f", &(ptr+i)->c, &(ptr+i)->a, &(ptr+i)->b); 
  }
}

Upvotes: 1

Views: 281

Answers (4)

AnT stands with Russia
AnT stands with Russia

Reputation: 320361

Firstly, operator & in C and C++ languages is used to get address of any variable. More precisely, it can be used to get address of [almost] any lvalue (there are some differences between C and C++ in that regard). Pointer variables are ordinary variables. There's nothing special about them, which means that there's nothing unusual in seeing operator & applied to pointers.

Secondly, in the code you provided, there actually isn't a single instance of & being applied to a pointer. There are four applications of & in your code

&n
&(ptr+i)->a
&(ptr+i)->b
&(ptr+i)->c

In the first two cases it is applied to int objects. In the thirds case it is applied to float object. In that last case it is applied to a char [30] object. No pointers.

Upvotes: 1

bstamour
bstamour

Reputation: 7766

The code &(ptr + i)->c gives you the address of the struct element c that belongs to the ith struct in your list. Let's break it down a bit.

(ptr + i) is pointer arithmetic. It adds i to the address stored at ptr.

(ptr + i)->c accesses the struct element c through the pointer (ptr + i).

&(ptr + i)->c takes the address of the struct element c through the pointer (ptr + i).

Also, I know this isn't quite doing what you thought it was doing, since you thought the address-of operator applied to the pointer, but just an FYI: you can indeed take the address of a pointer. Such a construct is a pointer to a pointer, and is useful when you want to change the pointer (not just the value stored at the address it points to) in a function. e.g.

int a = 5; /* regular variable */
int* pa = &a; /* pointer to a */
int** ppa = &pa; /* pointer to pointer (which points to a) */

Upvotes: 2

Zac Howland
Zac Howland

Reputation: 15872

struct name* ptr;

This creates a pointer to a name.

struct name** p = &ptr;

This creates a pointer to a pointer to a name by taking the address of the pointer you already created.

In your case, you are passing in pointers to scanf, and have a dynamic array of name, so 

&(ptr + i)->c

finds the ith element of the array, and returns the address of its c member to scanf (same with a and b).

Passing in the address of the pointer allows for that pointer to be changed (e.g. reallocated). In C++, it would be virtually identical to passing by reference.

Upvotes: 0

pippin1289
pippin1289

Reputation: 4939

&(ptr + i)->c

this takes the address of the variable c stored at the ith element of ptr. So your first intuition was correct.

ptr+i

is simply the pointer arithmatic to find the ith element fo the array.

(ptr+i)->c

accesses the field c from that struct, and the & takes the address of that variable.

Upvotes: 2

Related Questions