C Programming: Analyzing Array Pointer Arithmetic and Output

"#include <stdio.h>\n\nint main() {\nshort arr[3][2] = { 0x19, 2, 6, 4, 0xfff5, 3 };\nshort (p)[2] = (short ()[2])(&arr + 1);\n\nprintf("%d\n", (int)(*(arr + 2) + 2) - (int)(*arr + 2));\nprintf("%d\n", ((arr + 1) - *(*arr + 1)));\nprintf("%d\n", *(*p - 1));\nprintf("%d\n", *p[-1]);\nprintf("%x\n", *(int *)((int)arr + 1));\n\nreturn 0;\n}\n\nThe output of the code is as follows:\n\n1. The line printf("%d\n", (int)(*(arr + 2) + 2) - (int)(*arr + 2)); prints the value 4. \n\nExplanation: \n- *(arr + 2) is equivalent to arr[2], which gives us the third element of the array arr. In this case, it is the array {0xfff5, 3}.\n- (int)(*(arr + 2) + 2) adds 2 to each element of the array, resulting in {0xfff7, 5}.\n- (int)(*arr + 2) adds 2 to the first element of the array arr, resulting in 0x1b.\n- Subtracting (int)(*arr + 2) from (int)(*(arr + 2) + 2) gives us the difference between the two arrays, which is 4. \n\n2. The line printf("%d\n", *(*(arr + 1) - *(*arr + 1))); prints the value 2. \n\nExplanation:\n- *(*arr + 1) is equivalent to *(*arr) + 1, which gives us the value 0x1a.\n- (arr + 1) gives us the second element of the array arr, which is the array {6, 4}.\n- *(arr + 1) - *(*arr + 1) subtracts 0x1a from each element of the array {6, 4}, resulting in the array {4, 2}.\n- *(*(arr + 1) - *(*arr + 1)) gives us the first element of the resulting array, which is 2. \n\n3. The line printf("%d\n", *(*p - 1)); prints the value 6. \n\nExplanation:\n- p is a pointer to the array {0xfff5, 3} in arr.\n- *p gives us the array {0xfff5, 3}.\n- *(*p - 1) gives us the first element of the array {0xfff5, 3}, which is 0xfff5. \n\n4. The line printf("%d\n", *p[-1]); prints the value 6. \n\nExplanation:\n- p[-1] gives us the array {6, 4} in arr.\n- *p[-1] gives us the first element of the array {6, 4}, which is 6. \n\n5. The line printf("%x\n", *(int *)((int)arr + 1)); prints the hexadecimal value 19. \n\nExplanation:\n- (int)arr gives us the memory address of the first element of the array arr.\n- (int)arr + 1 increases the memory address by 1 byte.\n- (int *)((int)arr + 1) casts the memory address to a pointer to an integer.\n- *(int *)((int)arr + 1) dereferences the pointer and gives us the value at that memory address, which is 0x19 in hexadecimal.