C Programming - Pointers - Discussion

@Bins Emmanuel.

Nice explanation, Thanks.

["IndiabiX"].

0-I
1-n
2-d
3-i
4-a
5-b
6-i
7-X
7["IndiabiX"]= it prints X.

Thanks everyone for explaining.

Thank you all for explaining it.

int main()
{
printf("%c\n", 7["IndiaBIX"]);//output=X. 7 is index. I=0, n=1,...X=7
printf("%c\n", "IndiaBIX"[7]);//output=X. both printf are same
printf("%d\n", 7["IndiaBIX"]);//output=88. prints ASCII value of X
printf("%c\n", *("IndiaBIX"+7));//output=X
printf("%c\n", *(7+"IndiaBIX"));//output=X
return 0;
}


For more details:

If "a" is an array and "i" is index.
a[i]<=>[i]a<=>*(a+i)<=>*(i+a) // all are same and in problem they have used 2nd case.

@Rhi.

a [i] translates to *(a+i) but you have declared it as a array and i is its size.

So, if you declare i as an array and a its size i,e, , i [a]
Then only you can say that but they are not equal definitely like you have explained [*(a+i)=*(i+a)].

Well in C, a[i] always gets translated to *(a + i)
So *(a + i) = *(i+ a)
a[i] = i[a].

Thanks for the explanation @Shabana.

Is it work for multi-dimensional array and what is syntax?

Please explain me.

The string "IndiaBIX" returns its base address.

Let us assume base address as 1000.

That is, an array name 'arr' also holds the base address of array arr.
We know that arr[7] == 7[arr] == * (arr + (7 * size of character 1 byte)).

Then in the above example,

7["IndiaBIX"] is calculated as below:

* (1000 + (7 * 1)) //where 1 is the size of char.

=> * (1007).

=> character stored in the location 1007.

=> X.

Hence X is the output.