Question

create "bitty.c" that uses bitwise operators and return statements to implement the functions in "bitty.h".

"bitty.c" that can only implement "bitty.h"

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bitty.h:

// Returns 1 if v is an even number, otherwise returns 0

unsigned char isEven(unsigned char v);

// Returns 1 if v is zero, otherwise returns 0

unsigned char isZero(unsigned char v);

// Returns 1 if a and b have the same value, otherwise 0

unsigned char equals(unsigned int a, unsigned int b);

// Returns a if a is greater than or equal to b, otherwise returns b

// You may use if but may not use comparison operators or else

// e.g. if(a) is allowed but if(a<b) and if(b==0) are not allowed

unsigned char geq(unsigned char a, unsigned char b);

// Returns the number of bits that are 1

unsigned char tally(unsigned int n);

Answer 1

If you understand the concept do give it a like:)

unsigned char isEven(unsigned char v){

if(v & 1)

return 0;

return 1;

}

Explanation:  If v is even then rightmost bit must be equal to zero, so after anding it with 1 it will always give answer as 0

unsigned char isZero(unsigned char v){

if(v | 1)

return 0;

return 1;

}

Explanation:  If v is zero then ORing it with 1 will always give answer as 0.

unsigned char equals(unsigned int a, unsigned int b){

if(a ^ b)

return 0;

return 1;

}

Explanation:  XOR of 2 same numbers is always zero as XOR of same digit is 0.

unsigned char geq(unsigned char a, unsigned char b){

if(!(a ^ b))

return a;  

if(!(b ^ 0) || (a / b)))){

return a;

}

return b;

}

Explanation:  Here a ^ b refers to a == b , !(b ^ 0) refers to b == 0 and a/b will give us quotient as 0 if a<b or any other number if a>=b .

So, a ^ b gives us zero that means they are equal so we will return a

else we will check that if b == 0 OR a / b> 0 then we will return a other wise we will return b.

unsigned char tally(unsigned int n){

int count = 0;

while(n){

if(n & 1)

count++;

n>>=1;  

}

return count;

}

Explanation: Here we are checking each bit by Anding it with 1 if we encounter 1 then we will update our counter.

I hope I am able to solve your problem, if yes then do give it a thumps up :)