How do sizeof(arr) / sizeof(arr[0]) work?

Question

When looking for a size of an array in a for loop I've seen people write

int arr[10];
for(int i = 0; i < sizeof(arr) / sizeof(arr[0]); i++){}

How is sizeof(arr) / sizeof(arr[0]) the length of the array? How does it technically work?

Answer 1

If you have an array then sizeof(array) returns the number of bytes the array occupies. Since each element can take more than 1 byte of space, you have to divide the result with the size of one element (sizeof(array[0])). This gives you number of elements in the array.

Example:

std::uint32_t array[10];

auto sizeOfInt = sizeof(std::uint32_t); // 4
auto numOfBytes = sizeof(array); // 10*sizeOfInt = 40
auto sizeOfElement = sizeof(array[0]); // sizeOfInt = 4
auto numOfElements = sizeof(array) / sizeof(array[0]); // numOfBytes / sizeOfElement = 40 / 4 = 10

LIVE EXAMPLE

Note that if you pass an array to a function, the above won't work since the array decays to a pointer and sizeof(array) returns the size of the pointer.

std::size_t function(std::uint32_t a[]) // same for void function(std::uint32_t a[10])
{
    return sizeof(a); // sizeof(std::uint32_t*)!
}

std::uint32_t array[10];
auto sizeOfArray = function(array); // array decays to a pointer inside function()

LIVE EXAMPLE #2