Thread: Advanced task, implementing a function :]

Have you made an attempt to do this?
Are you familiar with bitwise operators?

Originally Posted by Matticus

Have you made an attempt to do this?
Are you familiar with bitwise operators?

Hi, well yes.
I haven't written it yet but I have an idea.
I know I should first ensure that the two exponents are equal. If not, I can take the higher one and decrease (or take the lower one and increase), but then I need to change the Mantissa, like shifting it, but Im not sure how exactly to do this in C and what exactly I need to change.
Yes, Im familiar with bitwise operations, but I think I have a way to go before I get to this point in the function.
Can you help me figuring out how should I continue from here? Do I must using masking somehow?

Thanks

anyone?

So I wrote this code. But unfortunately it doesnt work with my checks. I need you to help me understand what Im missing here:

Code:

unsigned long add(unsigned long float1, unsigned long float2) {
	char E1=float1>>23, E2=float2>>23, E3;
	int M1, M2, M3, M_mask=0x1, ofCheck=(0x1)<<24, shift, res=0x0;
	M_mask=(M_mask<<24)-1;
	M1=M_mask&float1;
	M2=M_mask&float2;
	
	if (E1<E2) {							//aligning exponents and normalizing the Mantissas
		E1=E2;
		shift=E2-E1;
		M1>>=shift;
	}
	else if (E2<E1) {
		E2=E1;
		shift=E1-E2;
		M2>>=shift;
	}


	M3=M1+M2;
	if (!(M3&ofCheck))						//checking overflow(24th bit) in mantissa
		E3=E1;
	else {
		M3>>=1;
		E3+=1;
	}
	
	res|=E3;							//building the result
	res<<=31;
	res+=M3;


	return res;
	
}

Here is one problem:

Code:

	E1=E2;
	shift=E2-E1;

"shift" will always = 0

-

Originally Posted by megafiddle

Here is one problem:

Code:

    E1=E2;
    shift=E2-E1;

"shift" will always = 0

-

Hi.
Well I fixed it and it still returns wrong numbers. My main is this:

Code:

#include <stdio.h>


unsigned long add(unsigned long float1, unsigned long float2);


int main () {
	unsigned long f1=6.5L, f2=0.6L, result;
	result = add(f1,f2);
	printf("%lu", result);
	return 0;
}


unsigned long add(unsigned long float1, unsigned long float2) {
	char E1=float1>>23, E2=float2>>23, E3=0;
	unsigned long M1, M2, M3, M_mask=0x1, ofCheck=(0x1)<<24, res=0x0;
	int shift;
	M_mask=(M_mask<<24)-1;
	M1=M_mask&float1;
	M2=M_mask&float2;
	if (E1<E2) {										//aligning exponents and normalizing the Mantissas
		shift=E2-E1;
		E1=E2;
		M1>>=shift;
	}
	else if (E2<E1) {
		shift=E1-E2;
		E2=E1;
		M2>>=shift;
	}


	M3=M1+M2;
	if (!(M3&ofCheck))									//checking overflow(24th bit) in mantissa
		E3=E1;
	else {
		M3>>=1;
		E3+=1;
	}
	
	res|=E3;											//building the result
	res<<=31;
	res+=M3;


	return res;
	
}

I would be glad if you or someone else can debug or run it and see what I mean.
Thanks

Originally Posted by Salem

Perhaps your trouble starts with this.
unsigned long f1=6.5L, f2=0.6L, result;

You end up with 6 and 0 respectively.

Not the bit-pattern of floating point numbers mysteriously packed into an unsigned long.

Do you have any clue what I should assign there instead?

Originally Posted by laserlight

You might want to write a function that takes an argument of type double and converts it to an unsigned long whose bits correspond to that double's value according to the rules laid out in your assignment.

Can you write a short example for how to do this?

Originally Posted by grumpy

I'm reasonably certain laserlight could. However, her point is that YOU need to try doing it. You will learn more by attempting to do it yourself than you will if someone else does it for you.

I agree. I just didnt fully understand what she meant.. I didnt understand what that function she has suggested should do. Thats why Ive asked for a short example. Trust me I dont want anybody to do the job for me... So, can you or anyone else try and calrify this to me and explain me what I should do to fix my problem please?

Thanks