Using addressof with memcpy of 2 arrays

Question

Why is it that both the memcpy below produces the same result:

int arr1[5] = {some_data1};
int arr2[5] = {some_data2};

memcpy(&arr1, &arr2, (5 * sizeof(int)));
memcpy(arr1, arr2, (5 * sizeof(int)));

Why doesn't the first memcpy see a double pointer for arr1 and arr2? Is it because the array name is not truly a pointer but rather an identifier for a variable of the "array" type?

possible duplicate of [How come an array's address is equal to its value in C?](http://stackoverflow.com/questions/2528318/how-come-an-arrays-address-is-equal-to-its-value-in-c) — Barmar, Jun 13 '14 at 12:40

Sergey Kalinichenko · Answer 1 · 2014-06-13T12:53:19.513

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Why doesn't the first memcpy see a double pointer for arr1 and arr2?

Bluntly speaking, because it is not a double pointer.

Is it because the array name is not truly a pointer but rather an identifier for a variable of the "array" type?

Absolutely! This happens precisely because arrays are not pointers, despite many misleading similarities shared by the two kinds of C objects.

The behavior is the same because both expressions produce numerically identical void* pointers when passed to memcpy.

edited Jun 13 '14 at 12:53

answered Jun 13 '14 at 12:40

Sergey Kalinichenko

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Thank you. Updated my question. Hasty pseudo code, I apologize. – Stas Bukhtiyarov Jun 13 '14 at 12:51

score 1 · Answer 2 · answered Jun 13 '14 at 12:39

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Because in C, when a is an array we have this:

 (void *) a == (void *) &a

the address of &a (pointer to the array) and of &a[0] (pointer to the first element of the array) is the same, only the type is different.

answered Jun 13 '14 at 12:39

ouah

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score 0 · Answer 3 · edited May 23 '17 at 12:13

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Duplicate of this question - "The name of an array usually evaluates to the address of the first element of the array, so array and &array have the same value."

edited May 23 '17 at 12:13

Community

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answered Jun 13 '14 at 12:39

jarmod

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If it's a duplicate, use the _flag_ button, don't post an answer that just points to the other question. – Barmar Jun 13 '14 at 12:40
@jarmod Thank you, I didn't find that question prior to posting. Apologies. – Stas Bukhtiyarov Jun 13 '14 at 12:43

score 0 · Answer 4 · answered Jun 13 '14 at 12:40

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In some contexts, when you use an array name it is converted to the address of the array. Passing an array as a function argument is one of those contexts. But using the array with the & operator is not -- it returns the address of the array.

answered Jun 13 '14 at 12:40

Barmar

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score 0 · Answer 5 · answered Jun 13 '14 at 12:51

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in array, your array variable(arr1 amd arr2) naturally represent base address of the array, you do not need to give &(address of operator). arr1==&arr1

answered Jun 13 '14 at 12:51

user3614789

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Using addressof with memcpy of 2 arrays

5 Answers5