ARR is a metric used to measure the total revenue that a SaaS company expects to generate from its subscriptions. Annual recurring revenue (ARR) is a metric used to measure the total revenue that a ... As others have mentioned, *(&arr + 1) triggers undefined behavior because &arr + 1 is a pointer to one-past-the end of an array of type int [7] and that pointer is subsequently dereferenced.
How *(&arr + 1) - arr is working to give the array size Here arr is itself a pointer to the type char* Note: In case 1 array arr degrades to a pointer to become the type char** but it's not possible the other way around i.e, pointer in case 2 cannot become an array. 7 what does *arr[] mean? As standalone expression *arr[] is not valid. For variable definitions there are two meanings here, depending of the context in which such an expression appears: Variable definition with initialiser (as per the OP's snippet) Copy When you use arr in your function call, it will decay to a pointer to its first element, it's equal to &arr[0].
arr calculation, It will have the type int*. If you use arr[i] (for any valid index i), then you pass the value of a single element of the array. It will have the type int. Since the function expects a pointer (when declaring an argument, int arr[] is equal to int* arr), passing plain arr will work ... arrays - C++: What is the difference between ' arr ' and ' arr ...
arr calculation, In fact it is just interpreting that it needs to go till the boundary as arr [::1] gives normal array. Is this just coded as a special case or is there something more going on?
