Say: arr.reduce((bestIndexSoFar, currentlyTestedValue, currentlyTestedIndex, array) => currentlyTestedValue > array[bestIndexSoFar] ? currentlyTestedIndex : bestIndexSoFar, 0);, which can be described as: iterate the array starting from index 0 (2nd parameter), if currentlyTestedValue is higher than the value of the element at the ...
int arr[3]; cout << arr << endl; cout << &arr << endl; Remark: This question was closed, but now it is opened again. (Thanks ?) I know that &arr[0] and arr evaluates to the same number, but that is not my question! The question is why &arr and arr evaluates to the same number. If arr is a literal (not stored anyware), then the compiler should complain and say that arr is not an lvalue. If the ...
Your question involves a mix of basic Python syntax, and numpy specific details. In many ways it is the same for lists, but not exactly. arr[:, 0] returns the 1st column of arr (a view), arr[:,0]=10 sets the values of that column to 10. arr[:] returns arr (alist[:] returns a copy of a list). arr[:]=arr2 performs an inplace replacement; changing the values of arr to the values of arr2. The ...
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
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?
&arr (I think this point to the address of the first element of arr) *(&arr) then I don't quite understand what this do, what does the symbol * do to &arr (i.e. to the address here)?, because the two outputs are the same when I run them and finally what is it exactly happening when an integer say 1 is added to the address by this code here ...
How does *(&arr + 1) - arr give the length in elements of array arr?