Different Pointer Arithmetic Results when Taking Address of Array

Question

Program:

#include

int main(void) {
    int x[4];
    printf(\"%p\\n\", x);
    printf(\"%p\\n\", x + 1);
    printf(\"%p\\n\", &x);
    p         


        

Answer 1

Even though x and &x evaluate to the same pointer value, they are different types. Type of x after it decays to a pointer is int* whereas type of &x is int (*)[4].

sizeof(x) is sizeof(int)*4.

Hence the numerical difference between &x and &x + 1 is sizeof(int)*4.

It can be better visualized using a 2D array. Let's say you have:

int array[2][4];

The memory layout for array is:

array
|
+---+---+---+---+---+---+---+---+
|   |   |   |   |   |   |   |   |
+---+---+---+---+---+---+---+---+

array[0]        array[1]
|               |
+---+---+---+---+---+---+---+---+
|   |   |   |   |   |   |   |   |
+---+---+---+---+---+---+---+---+

If you use a pointer to such an array,

int (*ptr)[4] = array;

and look at the memory through the pointer, it looks like:

ptr             ptr+1
|               |
+---+---+---+---+---+---+---+---+
|   |   |   |   |   |   |   |   |
+---+---+---+---+---+---+---+---+

As you can see, the difference between ptr and ptr+1 is sizeof(int)*4. That analogy applies to the difference between &x and &x + 1 in your code.