Question

PLEASE ANSWER I WILL RATE YOUR ANSWER AND THUMBS UP

For the following C functions:

int q3(int x) {

    return ((~x)>>31) & 0x1;

}

int q4(int x) {

    int mask = x>>31;

    int y= (x ^ mask);

    int z = (~mask + 1);

    return (y+z);

}

int q5(int x){

    int y = !x;

    int z = x >> 31;

    z = z | y;

    return !z;

}

int q6(int x) {

    int result = (x<<31);

    result = (result >> 31);

    return result;

}

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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

IMP NOTE: int in c is of 4 bytes, ie 32 bits. The most significat bit (left most) will be 1 for -ve numbers and 0 for positive numbers.

Function q3 will return 1 for positive numbers or 0 and 0 for negative numbers

~x will compliment the number and >>31 will bring the MSB to the LSB and anding with 1 will return 1 for positive numbers and 0 for -ve numbers

Here is the simpler version:

int newq3(int x)
{
    if (x < 0)
        return 0;
    return 1;
}

q4 will always return the absolute value of the given number

mask will store the x shifter to 31 times so 1 for -ve numbers and 0 for +ve

y will be 0 for even -ve numbers and 1 for odd -ve numbers, and always 0 for +ve numbers

Then we complement the mask and add 1 to get the original number by in +ve format

Here is simpler version

int newq4(int x)
{
    if (x < 0)
        return x * (-1);
    return x;
}

q5 will return 1 for +ve numbers and 0 for -ve numbers and 0

y becomes 1 for 0 and 0 for -ve and +ve numbers

z become the msb digits of the numbers

then z = z|b becomes 1 if z or y is 1 ( for 0 or -ve numbers) and 0 for +ve numbers

!z becomes reverse of z

Here is the simpler version

int newq5(int x)
{
    if (x <= 0)
        return 0;
    return 1;
}

q6 returns -1 for odd numbers and 0 for even numbers

For even numbers the result clearly becomes 0 after shifting so it will remain 0 after another shift too

For odd numbers the lsb is always 1, so after left shifting 31 bits result will have 1 at the MSB and after right shifting 31 times 1 will come at lsb and sign will be preserved resulting in -1

Here is the simpler code:

int newq6(int x)
{
    if (x < 0)
        x = x * (-1);
    if (x % 2 == 1)
        return -1;
    return 0;
}