low-level

I understand that arrays in C are allocated in row-major order. Therefore, for a 2 x 3 array: 0 1 2 3 4 5 Is stored in memory as 0 1 2 3 4 5 However, what if I have a 2 x 3 x 2 array: 0 1 2 3 4 5 and 6 7 8 9 10 11 How are these stored in memory? Is just consecutive like: 0 1 2 3 4 5 6 7 8 9 10 11 Or is it some other way? Or does it depend on something? All "dimensions" are stored consecutively in memory. Consider int arr[4][100][20]; you can say that arr[1] and arr[2] (of type int[100][20] ) are contiguous or that arr[1][42] and arr[1][43] (of type int[20] ) are contiguous or that arr[1][42][7

A friend and I are going back and forth with brain-teasers and I have no idea how to solve this one. My assumption is that it's possible with some bitwise operators, but not sure. CMS In C, with bitwise operators: #include<stdio.h> int add(int x, int y) { int a, b; do { a = x & y; b = x ^ y; x = a << 1; y = b; } while (a); return b; } int main( void ){ printf( "2 + 3 = %d", add(2,3)); return 0; } XOR ( x ^ y ) is addition without carry. (x & y) is the carry-out from each bit. (x & y) << 1 is the carry-in to each bit. The loop keeps adding the carries until the carry is zero for all bits. ackb

In the Javadoc for Object.hashCode() it states As much as is reasonably practical, the hashCode method defined by class Object does return distinct integers for distinct objects. (This is typically implemented by converting the internal address of the object into an integer, but this implementation technique is not required by the Java™ programming language.) It's a common miconception this has something to do with the memory address but it doesn't as that can change without notice and the hashCode() does not and must not change for an object. @Neet Provided a link to a good answer https:/