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Here I have a pointer ptr to array arr of 4 integers. ptr points to the whole array. ptr[0] or *ptr points to the first element of the array, so adding 1 to ptr[0] gives the address of the second element of the array.

I can't understand why using sizeof(ptr[0]) gives the size of the whole array, 16 bytes, not the size of only the first element, 4 bytes, (as ptr[0] points to the first element in the array).

int arr[4] = {0, 1, 2, 3};
int (*ptr)[4] = &arr;
printf("%zd", sizeof(ptr[0])); //output is 16
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  • Shouldn't the second line be int *ptr = arr; ? That would make it point to the start (first element of) the array, which is equivalent to &arr[0]. Commented Oct 27, 2019 at 16:48
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    @AndreasWenzel Shouldn't the second line be int *ptr = arr;? Actually, no. int (*ptr)[4] creates ptr as a pointer to a full array of four int values. Pointer syntax like that is necessary to dynamically allocate true-multidimensional arrays. The "2-dimensional arrays" created with nested malloc() loops and wrongly described as multidimensional arrays are really 1-d arrays of pointers to multiple 1-d arrays. See stackoverflow.com/questions/42094465/… Commented Oct 27, 2019 at 17:02

4 Answers 4

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OP: ptr[0] points to the first element in the array.

Type confusion. ptr[0] is an array.

ptr is a pointer to array 4 of int.
ptr[0], like *ptr deferences the pointer to an array.
sizeof(ptr[0]) is the size of an array.

With sizeof(ptr[0]), ptr[0] does not incur "an expression with type ‘‘pointer to type’’ that points to the initial element of the array object" conversion. (c11dr §6.3.2.1 3). With sizeof, ptr[0] is an array.

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17 Comments

@Abd-ElrahmanMohamed Agree "But ptr[0][0] is an integer not points to integer ". "ptr[0] is the address of first element in the array" is not true.
@Abd-ElrahmanMohamed yes, the values are same but types are different. ptr[0] has array type and &ptr[0][0] has int * type
@chux ptr[0] (implicitly converted to int *) would evaluate to address of the first int element.
@chux about sizeof - right, I misunderstood the context of what you said
@Abd-ElrahmanMohamed That does not. printf("someforamt", ptr[0] , ptr[0]+1) does something different than sizeof(ptr[0]). The ptr[0] in the first case goes through an inplicit conversion. With sizeof(ptr[0]), ptr[0] does not.
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ptr here is of type pointer to an array of 4 int elements and the array type has size 16 on your platform (sizeof(int) * (number of elemetns)).

I can't understand why using sizeof(ptr[0]) gives the size of the whole array 16 bytes not the size of only the first element 4 bytes

because C type system has array types. Here both arr and *ptr has it. What you declare that you have. To get sizeof int here you should sizeof(ptr[0][0]) - where ptr[0] evaluates to array.

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with int (*ptr)[4] = &arr ; you have a pointer to an array of four integers and pointing to arr.

ptr is now pointing to arr, like a double pointer. We can access elements of arr using ptr[0][x] where x could be 0 to 4.

So sizeof(ptr[0]) is same as sizeof(arr)

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By definition, ptr[0] is the same as *(ptr + 0) which in turn is the same as *ptr. Further, ptr is initialized with &arr, so *ptr is *&arr and that is just arr. Note that the intermediate storage of &arr in ptr does not perform any array decay, so the equivalence is maintained and no type information is lost.

Note that all this is computed at compile-time, just to avoid this additional pitfall.

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