CODING:
#include< stdio.h >
int main()
{
char a[]="INISROH";
printf("%c",4[a]);
return 0;
}
OUTPUT
R
How?
a[i]=*(a+i) (we can represent a[i] using the ptr like this)
*(a+i)=*(i+a)
*(i+a)=i[a]
here 'i' is 0,1,2,...n
n is length of string
Number[character array]
Today i'm going to give you a interesting logical programs.
CODING:
#include< stdio.h >
int main()
{
char a[]="INISROH";
printf("%c",4[a]);
return 0;
}
R
a[i]=*(a+i) (we can represent a[i] using the ptr like this)
*(a+i)=*(i+a)
*(i+a)=i[a]
here 'i' is 0,1,2,...n
n is length of string
CODING:
#include< stdio.h >
int main()
{
char a[]="INISROH";
printf("%c",4[a]);
return 0;
}
R
a[i]=*(a+i) (we can represent a[i] using the ptr like this)
*(a+i)=*(i+a)
*(i+a)=i[a]
here 'i' is 0,1,2,...n
n is length of string
