Programming 2024 test series

Question

Caption

@sachinmittal1 sir here answer should be B na because Arr[2] point s to the 0th index of 2nd 1D-array so it means pointing to address of arr[2][0] so address size would be  8.

Comments

@sachinmittal sir ?

---

@JayRathi , bro arr[2]  firstly , arr points to 1D array of size 4. then arr+2 points to 3^rd row now *(arr+2) is an array of 4 int elements we can also write arr[2] . so sizeof this is sizeof(int)*no. of elements i.e 16.   

Answer 1

 

int arr[3][4];: A 2D array with 3 rows and 4 columns, totaling 12 integers.

int (*p)[12];: A pointer to an array of 12 integers.

p = malloc(sizeof(int) * 24);: Allocates memory for 24 integers (96 bytes, sizeof(int) = 4).

---

 

sizeof(arr[2]): arr[2] refers to the third row of the 2D array arr, which is an array of 4 integers. sizeof(arr[2]) will be the size of this array, 4 * sizeof(int) = 4 * 4 = 16 bytes.

sizeof(p): p is a pointer, and the size of a pointer is 8 bytes (given sizeof(int *) = 8).

sizeof(*p): *p dereferences the pointer p, which points to an array of 12 integers. sizeof(*p) will be the size of this array, 12 * sizeof(int) = 12 * 4 = 48 bytes.