Question

PLEASE SOLVE I WILL THUMBS UP AND RATE YOUR ANSWER WELL

For the following two C functions:

int q1(int x) {

int m8 = 0x55;

int m16 = m8 | m8 << 8;

int m32 = m16 | m16 <<16;

int z = x | m32;

return !(~z);

}

int q2(int x) {

int m8 = 0x55;

int m16 = m8 | m8 << 8;

int m32 = m16 | m16 <<16;

int z = x & m32;

return !!z;

}

Part i) Explain what each function does without running the code. EXPLAIN YOUR ANSWER OR YOU WILL NOT RECEIVE CREDIT.

Part ii) For each of the above functions, write a simpler version of the function (if there is one)

Answer 1

So talking about first function q1 so it initializes m8 = 0x55

m8 = 00000000 01011000

m8 << 8 = 01011000 00000000 //shifts left by 8 times

now m16 = m8 | m8<<8 //Bitwise OR of m8 and m8<<8

that will be m16 = 01011000 01011000

similarly

m16 << 16 = 01011000 01011000 00000000 00000000

so m32 = m16 | m16 << 16

m32 =  01011000 01011000 01011000 01011000

now z = x | m32 // x is passed from argument is done Bitwise OR with m32

return !(~z) //so here basically '~' toggles every bit of z and '!' is not so atlast it return 0

now in function q2 m32 is similarly calculated as above

m32 =  01011000 01011000 01011000 01011000

now z = x & m32;

now x is done Bitwise AND with m32 stored in z

and return !!z not of not will be 1 if z is any number else than 0 if z is zero then not of not of 0 will be 0

now coming to your second question any simpler version i think this is simpler nothing complex