IEEE 754 - signbit, expBits, fractbits,normalized

I am taking in a 8 digit hexadecimal number as an IEEE 754 bit floating point number
and i want to print information about that number( signbit, expbits, fractbits, normalized,
denormalized, infinity, zero, NAN) floating point should be a single.


I read up on bit shifting, and i think this is how i am suppose to do it?. however, i am not 100% sure. I understand that the sign bit is found in the left most position of the number. which indicates positive or negative. How much do i shift it to find each? do i just keep shifting it to find each one? Can someone explain how i am to find each one?

would i shift by 1 to find the signbit?
would i shift by 8 to get the exponent?
would i shift by 23 to get the frac?

if so how do i test it after i shift?

this is what i have so far

Code:

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char *argv[])
{


	int SwapTest[2] = { 1, 0 };
	int HexNumber;

	printf("Hex IEEE - 754\n");


	if( *(short *) SwapTest == 1 )
	{
		//little endian
		printf("\nbyte order: little-endian\n");
	}
	else
	{
		printf("byte order: big-endian\n");
	}
	printf("\n>");
	scanf("%x", &HexNumber);
	printf("\n%#x",HexNumber);




	return 0;
}

My input would be
40000000

and i get
0x40000000
which is what i want..

Yeah, shifting the correct number of units right, and ANDing with the correct number of set bits is how you would normally do this. For example:

Code:

int eg = 0xEA;//1110 1010
(eg>>3)&0x7;

This will get you the 4th through 6th least significant digits, here resulting in the number 5.

I am still lost.

Originally Posted by mrsirpoopsalot

I am taking in a 8 digit hexadecimal number as an IEEE 754 bit floating point number

No you're not. You're just reading in an int as hex. There is no 'float' or 'double' anywhere in that program.

Surely you actually mean to input a float?

No, I think he wants to input as hex, and output what each part of the floating point variable would be if he reinterpret cast it to float.

That doesn't require having float anywhere in the program, since getting the parts of the number requires reinterpreting back to an integer type, and the two casts cancel each other out.

Here's spelling it out a bit more:

Code:

int eg = 0xEA;//1110 1010 
//desired number in green
//olive shows where 3 comes from
// 0x7 is 111b, which is the bitmask to show three digits.
(eg>>3)&0x7;

I've written an article that might help you, although it's not exactly what you want. I use doubles, though I list modifications to make it work for floats. Also, you should be able to modify it easily enough to use a hex representation: Displaying the Raw Fields of a Floating-Point Number - Exploring Binary

Originally Posted by King Mir

No, I think he wants to input as hex, and output what each part of the floating point variable would be if he reinterpret cast it to float.

That doesn't require having float anywhere in the program, since getting the parts of the number requires reinterpreting back to an integer type, and the two casts cancel each other out.

That'll make it really hard to ensure that the code is correct if the data is never viewed in float form at any point. The code could just as easily be disecting an arbitraryly made up floating point format.

King mir is correct -

No, I think he wants to input as hex, and output what each part of the floating point variable would be if he reinterpret cast it to float.

I am understanding this more. The IEEE single precision floating point standard representation requires a 32 bit word, which may be represented as numbered from 0 to 31, from 0 on the RIGHT to 31 on the LEFT.. The first bit is the sign bit, the next eight bits are the exponent bits, and the final 23 bits are the fraction.

Therefore, to extract the sign bit we use the appropriate mask and shift by 31 to get the sign at the end?

In addition to get the value of the exponent. since, its 8 bits in length, we shift by 31 - 8(23) to shift it to the end?

If true then the mantissa requires no shifting? Also, to extract the value we use the mask. This part confuses me.

I think Bit masks are used to access specific bits in a byte of data?
if so, How do we come up with that hex number(bit mask)?
How do we know how many digits to show?

@DoctorBinary - I love your webpage..

Okay I get that you want to input data as hex. But how the heck do you know what float number the random hex that you enter corresponds to?
How do you know that you typed in the correct data?
How will you even know that your program is producing the correct result?

DoctorBinary: Why don't use a compile-time assert to check (and self-document) your compile-time assumption? Then you don't have any runtime penalty and there is no reason to take it out. Heck you don't even have to run the program to check the assertion! In MSVC, just use C_ASSERT for this.

Originally Posted by mrsirpoopsalot

@DoctorBinary - I love your webpage..

Thanks.

The bitmasks are just the hex equivalent of binary. So 0x7FF equals 011111111111 in binary; that is, a leading 0 and eleven 1 bits. You set the bits to 1 or 0 depending on how you are masking -- with "and" or "or". (See Mask (computing) - Wikipedia, the free encyclopedia, for example, to learn more about bitmasks).

Originally Posted by iMalc

DoctorBinary: Why don't use a compile-time assert to check (and self-document) your compile-time assumption? Then you don't have any runtime penalty and there is no reason to take it out. Heck you don't even have to run the program to check the assertion! In MSVC, just use C_ASSERT for this.

Yes, that would be cleaner. Can you give me an example that is compiler independent?

can someone help me?
I am still stuck. I have found the sign bit, exponent bit, and mantissa. how can i tell if its normalized or denormalized? What about infinity? zero? or NAN
please i want to learn but i dont know what to do. Can someone give me clear instructions? or some working code?

This is what i have so far..

Code:

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char *argv[])
{


	int HexNumber;

	

   
 
   int a = 0x12345678;
   unsigned char *c = (unsigned char*)(&a);
   if (*c == 0x78)
   {
      printf("\nlittle-endian\n");
   }
   else
   {
      printf("\nbig-endian\n");
   }

	printf("\n>");
	scanf("%x", &HexNumber);
	printf("\n%#x",HexNumber);


   bool negative = !!(HexNumber & 0x80000000);
   int exponent = (HexNumber & 0x7f800000) >> 23;
   int mantissa = (HexNumber & 0x007FFFFF);

	
    printf("\nsignBit %d,", negative);
    printf("expbits %d,", exponent);
	printf("fractbits %#x,", mantissa);
 



	return 0;
}

Here is my output which is what i want..

Code:

little-endian

>c0000000

0xc0000000
signBit 1,expbits 128,fractbits 0,

fpclasify will tell you the answer to those things, but you'll need to cast it to a float for that, which is the third time I find myself saying that.

For the last time, reinterpret (i.e. cast) it as a float, to ensure that anything you're doing is actually correct at all.

iMalc,

For the last time, reinterpret (i.e. cast) it as a float, to ensure that anything you're doing is actually correct at all.

My code should accept user input in the form of 8 hexadecimal digits.
My code will interpret those 8 digits as an IEEE 754 32-bit floating point
number and will print out information about that number.

I want to use scanf to get the user input. My code should accept input with
or without "0x" in front of it (scanf does this).



Is this what you want me to do?

Code:

float HexNumber;

scanf("%x", &HexNumber);

Originally Posted by mrsirpoopsalot

can someone help me?
I am still stuck. I have found the sign bit, exponent bit, and mantissa. how can i tell if its normalized or denormalized? What about infinity? zero? or NAN

Single precision floating-point format - Wikipedia, the free encyclopedia

It explains the criteria for being normalized, denormalized, infinity, zero, -0, or NaN.

You would have to check which criteria fit.

Originally Posted by mrsirpoopsalot

My code should accept user input in the form of 8 hexadecimal digits.
My code will interpret those 8 digits as an IEEE 754 32-bit floating point
number and will print out information about that number.

I want to use scanf to get the user input. My code should accept input with
or without "0x" in front of it (scanf does this).

Your code is just saying "if this were the bits of a float, then the float would have e.g. an exponent of ...".

Is this what you want me to do?

Code:

float HexNumber;

scanf("%x", &HexNumber);

Not really. You don't have to change how you input the value. You already know how to treat the bytes as a different type because you did it here::

Code:

   unsigned char *c = (unsigned char*)(&a);

It's pretty similar to treat an int as a float:

Code:

   float f = *(float*)(&HexNumber);

Now you can do:

Code:

#include <float.h>

...

    if (_fpclass(f) == _FPCLASS_PD)
        printf("positive denormal!\n");
    if (_fpclass(f) == _FPCLASS_PINF)
        printf("positive infinity!\n");