hello guys,

here is my problem

#include<stdio.h>

main()

{

int a=2;

int b=9;

int c=1;

while(b)

{

if(odd(b))

c=c*a;

a=a*a;

b=b/2;

}

printf("%d\n",c);

}

How many times is c=c*a calculated?

Printable View

- 05-01-2002srinurocksc/c++ programming
hello guys,

here is my problem

#include<stdio.h>

main()

{

int a=2;

int b=9;

int c=1;

while(b)

{

if(odd(b))

c=c*a;

a=a*a;

b=b/2;

}

printf("%d\n",c);

}

How many times is c=c*a calculated? - 05-02-2002coug2197
Looks like C will be calculated twice. Since B is an int, and the assignment b=b/2 when b is origially 9 will result in 4.5, but because b is an int, the .5 is dropped off, and b is 4. And since 9 is odd, C is calculated and assigned. When b is 4, b is true and the while loop executes. It is not odd, however and does not caluclate C. When b=4/2, b is true, and b=2, which is even, thus C not calculated. After that b=2/2, b=1, which is true, and the while loop executes and since it 1 is odd, C is calculated again. Then b=1/2=.5, but just as was the case when b=9/2, b=4.5, the decimal part of an integer gets chopped off, and rounded downward. When b=1/2, it is rounded to 0 and this the while loop terminates. Hope this one helped.