Pointer doubt

Question

a and &a are the same thing where a is array name? I am unable to understand diff between  (a+1) and (&a+1) when assigned to a pointer

 

#include <stdio.h>

int main()

{

    int a[5] = {1,2,3,4,5};

    int *ptr = (int*)(&a+1);

    printf("%d %d", *(a+1), *(ptr-1));

    return 0;

}

Comments

a= what a contain i.e. address of array

&a= what is the address of a itself not what inside a.

2,5 is answer

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Here a is address of 1st element of an array.

&a is means address a pointer pointing to a. (address of an array)

when you increment &a by 1 due to address arithmetic pointer will now points to next element after last element! Because it is incremented by whole array size.

a+1 inside pointer will move on to 2nd element of an array. *(a+1) = value inside that address. = 2

Hence ans is 2,5

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@Ashwin Kulkarni When I run the following code both give me same output i.e a and &a print the same address..could u pls tell me why is it so?

#include<stdio.h>

int main(){

int a[2]={1,2};
printf("%d\n",a);
printf("%d",&a);
return 0;
}

 

Here the output I get is:

269667984
269667984