C Programming - Pointers - Discussion

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.

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.

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.

@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)].

It is just trick of mathematics. a[3] = 3[a] --> they are commutative both are expanded by compiler as *(a+3) and *(3+a) respectively. Even you can assign a variable to 3[a] = "X" which is same as a[3]="X".

Suppose take an example...


int a[3]={10,20,30};
printf("\n%d%d%d",0[a],1[a],2[a]);

same way as a[0] a[2]....
in the above example 7 is treated as index.

So it takes 7th character

Here in the program.
I - 0.
n - 1.
d - 2.
i - 3.
a - 4.
B - 5.
I - 6.
X - 7.

The compiler will skip all 6 character and it will 7th character X only because of 7["IndiaBIX"].

Hey program is good, because whenever complier can't define any data type or user define %c means char or you can modify this program or define %f,
%d then you will get satisfied answer.

printf("%c\n", 7["IndiaBIX"]);

It is printing the 7th character of the string "IndiaBIX".

I-0,
n-1,
d-2,
i-3,
a-4,
B-5,
I-6,
X-7.

If I write,

printf("%c,%c \n", "infocom"[5][3]);

It show error:

Subscripted value is neither array nor pointer.

Why this is some one can explain me?