CODEWITHSUNDEEP
carrays
Stas Bukhtiyarov
Stas Bukhtiyarov

Reputation: 97

Using addressof with memcpy of 2 arrays

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?

Upvotes: 1

Views: 134

Answers (5)

Sergey Kalinichenko
Sergey Kalinichenko

Reputation: 726967

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.

Upvotes: 3

user3614789
user3614789

Reputation: 200

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

Upvotes: 0

Barmar
Barmar

Reputation: 782166

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.

Upvotes: 0

jarmod
jarmod

Reputation: 78842

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."

Upvotes: 0

ouah
ouah

Reputation: 145899

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.

Upvotes: 1

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