CODEWITHSUNDEEP
csizeof
user1868357
user1868357

Reputation:

Sizeof arrays and pointers

Here is my sample code

#include<stdio.h>
void main()
{
 int arr[]={1,2,3,4,5,6};
 char *ptr,a;
 a='c';
 ptr=&a;
 int *ptr1,a1;
 a1=4;
 ptr1=&a1;
 printf("%d  %d   %d",sizeof(arr), sizeof(ptr1), sizeof(ptr));
}

Now, as far as I understand, size of will tell me the size required to store the variable, now the output for this one is

24 4 4

Why is the size of arr=24, after all it's just a pointer and it should be having size =4 ?

Thanks.

Upvotes: 2

Views: 433

Answers (5)

AnT stands with Russia
AnT stands with Russia

Reputation: 320719

"...after all it's just a pointer..."? No. Array is not a pointer. Array is an array object: a solid continuous block of memory that stores the array elements, no pointers of any kind involved. In your case array has 6 elements of size 4 each. That is why your sizeof evaluates to 24.

The common misconception about arrays being pointers has been debunked millions of times, but somehow it continues to pop up now and then. Read the FAQ, come back if you have any questions about it

http://c-faq.com/aryptr/index.html

P.S. As @Joachim Pileborg correctly noted in his answer, sizeof is not a function. It is an operator.

Another context in which arrays behave differently from pointers is the unary & operator (the "address of" operator). When unary & is applied to a pointer of type int * is produces a pointer of type int **. When unary & is applied to an array of type int [10] is produces a pointer of type int (*)[10]. These are two very different types.

int *p = 0;
int a[10] = { 0 };

int **p1 = &p;      /* OK */
int **p2 = &a;      /* ERROR */
int (*p3)[10] = &a; /* OK */

It is another popular source of questions (and errors): sometimes people expect & to produce a int ** pointer when applied to an int [10] array.

Upvotes: 5

Behnam Esmaili
Behnam Esmaili

Reputation: 5967

the sizeof() operator does not give you the number of elements in an array, it gives you the number of bytes a thing occupies in memory. Hence:

#include <stdio.h>

int main()
{
  char s[] = { 1, 2, 3, 4, 5, 0 };
  int xs[] = { 1, 2, 3, 4, 5, 0 };

  printf( "sizeof( s ) = %d\n",  sizeof( s  ) );
  printf( "sizeof( xs ) = %d\n", sizeof( xs ) );

  return 0;
}

sizeof( s ) = 6
sizeof( xs ) = 24

Upvotes: 0

koopajah
koopajah

Reputation: 25632

From Wikipedia

When sizeof is applied to the name of an array, the result is the size in bytes of the whole array. (This is one of the few exceptions to the rule that the name of an array is converted to a pointer to the first element of the array.)

Upvotes: 0

Joshua Green
Joshua Green

Reputation: 1575

Arrays don't decay into pointers inside sizeof; sizeof(arr) returns the total size allocated for the array which, in this case, is 6 * sizeof(int).

Upvotes: 0

Some programmer dude
Some programmer dude

Reputation: 409442

When you have an array, sizeof returns the size of the array (note that it's the size in bytes of the array, not the number of items in the array), otherwise it returns the size of the type or expression.

However, if you pass the array to a function, the array degrades to a pointer and the size information is lost.

Also, technically speaking sizeof is not a function, it's a special keyword in the language, and can be used without parentheses as well when used with expressions.

Upvotes: 0

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