C Programming - Pointers - Discussion

["IndiabiX"].

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

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.

Thanks everyone for explaining.

We can write a[i] = i[a].

Here they given second case,

Element[address]; i.e. When we are initializing a variable, it is mapping with address.

Eg: a ="indiabix".

a is stored with the base address of "indiabix" in code section.

So here,

7["indiabix"] = "indiabix"[7].

i.e assume address of indiabix stored in code section is 1000.

Then 1000[7].

1000[7] = *(1000+7) = *(1007) = x;

Note: The basic datatype of string is char, & ending with null char.

This follows the rule: arr[x] = *(arr + x) = *(x + arr) = x[arr] where, arr is an array and x is an integer value.

#include<stdio.h>

int main()
{
printf("%d\n", 7["IndiaBIX"]);
return 0;
}

If we write %d then output is 120 can any one explain me?

@Taseen: you're wrong.

Because 120 is the ASCII for lowercase x.
The ASCII for uppercase x is 88.
In your program it is uppercase so output should be 88 not 120.

If 0["indiabix"] then I.

If 1["indiabix"] then n and similarly others.

@Mj Taseen.

If %d is used it prints ASCII value of that character.

@Wikiok.

Nice ans/example.

Compiler will check condition and print given character.