Similarities and differences between arrays and pointers through a practical example

Question

Given the following code:

#include 
#include 

int main()
{
    int a[1];
    int * b = malloc(sizeof(int));

    /* 1 */
             


        

Answer 1

That's normal ...

First, scanf requires a pointer. "a" and "b" already are pointers ! So :

/* 1 */
scanf("%d", a);
printf("%d\n", a[0]);

/* 2 */ 
scanf("%d", b);
printf("%d\n", b[0]);

Will work.

Normally /* 1 */ shouldn't work. But gcc transforms "&a" by "a" because "&a" doesn't have any sense.

printf("&a = %p\n", &a);
printf("a = %p\n", a);
printf("&b = %p\n", &b);
printf("b = %p\n", b);

&a = 0x7ffff6be67d0
a = 0x7ffff6be67d0
&b = 0x7ffff6be67c8
b = 0xb0b010

You can't take the adress of a. But b is a "normal variable" of type pointer, and thus you can take it's address by "&b".

On /* 2 */ you're putting the value entered by the user in b and thus, *b (or b[0]) will crash unless the user will enter a valid readable memory address.