Expert: Narendra - 6/16/2004

Question
***can you explian the meaning for the follwoing program for me, I don't know what the meaning of the code.****


char *c1, *c2, *c3, *c4, *c5 ;
char analysis[8] = {'a', 'n', 'a', 'l', 'y', 's', 'i' ,'s'};

int main()
{
 c5 = c4 = analysis;
 ++c4;
 c3 = &analysis[6];
 c2 = c5 + 2 ;
 c1 = &analysis[7] - 3 ;

 printf ("%c\t%c\t%c\t%c\t%c", *c1,*c2,*c3,*c4,*c5);
 
 return 0;
}   
---------------------------------------------                  


y    a        i              n           a  

Answer
c1, c2, c3, c4 and c5 are pointers (which can hold addresses).
analysis is an array variable which is holding the string "analysis".
>c5 = c4 = analysis;
The starting address of the array is given to c5 and c4. Hence they point to first character 'a' in the array.
>++c4;
c4 is incremented by 1. Hence points to 'n' in the array.
>c3 = &analysis[6];
c3 is given the address of 7th character (count from 0) of the array. Hence c3 points to character 'i'.
>c2 = c5 + 2 ;
c5 points to first character. plus 2 means, c2 points to 3rd character 'a' (second 'a' in the string).
>c1 = &analysis[7] - 3 ;
c1 points to 3rd character from the end (not counting the last character).

c1 holds the address. *c1 means the data strored at c1. Since c1 points to 3rd character from the end (that is 5th character - count starts from 0), *c holds character 'y'.
Hence *c1 will print 'y' on the screen.
Similarly for other pointer variables *c2, *c3, *c4 and *c5.

Hope it is clear now.

-ssnkumar