CODEWITHSUNDEEP
carrayspointers
Ravi
Ravi

Reputation: 79

C pointers difference between accepting values using arrays and variables?

I'm new to pointers...

While assigning values to an array using pointer we use:

int *arr;
arr = malloc(10*sizeof(int));
for(i=0;i<10;i++)
{
   scanf("%d",(arr + i));
}

But while assigning to a variable we use

int *arr = malloc(sizeof(int));
*arr = 10;

So why cant we use,

scanf("%d",*(arr + i));

Why is it showing error?

Upvotes: 1

Views: 65

Answers (3)

user3078414
user3078414

Reputation: 1937

In 

scanf("%d",(arr + i));

(arr + i) already is a pointer to the i-th element of the array arr. Had you scanfed into a "single" variable, you would have had to pass its address (=pointer to that variable).

scanf("%d", &some_int_variable);

Why is it showing error?

Dereferencing:

scanf("%d",*(arr + i));

would be wrong, because it wouldn't pass the pointer to the i-th array member to read to, since scanf() requires a pointer to the argument specified by the format tag, not a value.

Upvotes: 2

AndersK
AndersK

Reputation: 36102

scanf takes an address (pointer) of a value, the format specifier tells what kind of pointer.

scanf("%d", (arr + i) ) means you are giving scanf the address arr plus an offset of i,

when you write *arr = 10; you can also write it as *(arr + 0) = 10; so you are de-referencing the value and assigning it 10 in other words you cannot give this to scanf since it wants a pointer.

so 

arr = malloc(10*sizeof(int));

      +---+---+---+---+---+---+---+---+---+---+
arr ->|0  |   |   |   |   |   |   |   |   | 9 |
      +---+---+---+---+---+---+---+---+---+---+

arr + i is an address in the range 0..9, de-referencing a value in the array you write *(a+i)

but 

arr = malloc(sizeof(int));

      +---+
arr ->|   |
      +---+

writing *arr you are de-referencing the one value you allocated

*arr = 10; 

      +---+
arr ->| 10|
      +---+

however writing 

scanf("%d", arr);

is fine since arr is a pointer that points to an integer and scanf takes a pointer.

Upvotes: 0

4386427
4386427

Reputation: 44329

Because scanf requires a pointer. Not a value. *(arr+i) is the value so if you passed that to scanf the function could not change the value so that it was effective after the function call.

It requires a pointer so that it can change the value of the object pointed to and make the change effective after the call.

It is like all function calls in c. They pass by value. Inside the function the value may be changed but as soon as the function returns any such changes are lost.

However, if the value passed is the value of a pointer, the function may change the value of the object pointed to by the pointer. Such a change will still be effective when the function returns.

A simple example:

void f(int a)
{
    a = 10;  // This change is lost when the function returns
}

void g(int* a)
{
     *a = 10;  // This change will survive when the function returns
}

int main(void)
{
    int a = 0;
    f(a);
    printf("%d\n", a); // will print 0
    g(&a);
    printf("%d\n", a); // will print 10
    return 0;
}

Upvotes: 0

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