Understanding arr[::-1] in numpy. Is this a special case? Is this just coded as a special case or is there something more going on? And is there a way to get reversed array view by explicitly specifying the three expressions in indexing i e , are there any values for expr1, expr2, expr3 which can reverse arr in arr[expr1:expr2:expr3]?
What is the meaning of arr [:] in assignment in numpy? arr=arr2 sets the object that the arr variable is pointing to Now arr and arr2 point to the same thing (whether array, list or anything else) arr[ ]=arr2 also works when copying all the data Play about with these actions in an interactive session Try variations in the shape of arr2 to see how values get broadcasted
What is the difference between * arr and * arr [0] in C++, if arr is an . . . 1 Suppose I have an array of integers called arr I am trying to understand the distinction between * arr and * arr[0] I read that in C++, arr is essentially a pointer to the first element in the array and arr is a pointer to the whole array They both return the same address, I get this part
c - What does *arr [] mean? - Stack Overflow 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) c Copy
Why is arr and arr the same? - Stack Overflow 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
Array increment positioning with respect to indexer in C - array [i . . . 3 41 4 5 In this example i = 3, so arr[3] = 40 It then increases the value from 40 to 41 So arr[i]++ increments the value of this particular index and a[i++] first increments the index and then gives the value for this index
How *( arr + 1) - arr is working to give the array size 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
Why does i[arr] work as well as arr[i] in C with larger data types? 0 arr is declared as the array of 4 structures with each structure comprising of a char array of 1024 chars arr is resolved as the pointer to the first element to this array of structures When you increment arr by 1, this new pointer, will skip one whole structure ( the type to which it points ) and then points to the next element in the array
c - Difference between *arr [] and **arr - Stack Overflow Theoretically, *arr[] and **arr are different For example : char *arr[size]; case 1 Here arr is a an array of size size whose elements are of the type char* Whereas, char **arr; case2 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