IEEE 754 representation

Question

Can someone look over my program and tell me if i am doing it correctly?

I am accepting user input in the form of 8 hexadecimal digits. I want to interpret those 8 digits as an IEEE 754 32-bit floating point number and will print out information about that number.

here is my output:

IEEE 754 32-bit floating point

byte order: little-endian

>7fffffff

0x7FFFFFFF
signBit 0, expbits 255, fractbits 0x007FFFFF
normalized:   exp = 128
SNaN

>40000000

0x40000000
signBit 0, expbits 128, fractbits 0x00000000
normalized:   exp = 1

>0

0x00000000
signBit 0, expbits 0, fractbits 0x00000000
+zero

here is the code..

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

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


int HexNumber;
int tru_exp =0;
int stored_exp;
int negative;
int exponent;
int mantissa;

printf("IEEE 754 32-bit floating point");


int a = 0x12345678;
unsigned char *c = (unsigned char*)(&a);
if (*c == 0x78)
{
    printf("\nbyte order: little-endian\n");
}
else
{
    printf("\nbyte order: big-endian\n");
}

do{

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

    printf("\n0x%08X",HexNumber);

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


printf("\nsignBit %d, ", negative);
printf("expbits %d, ", exponent);
printf("fractbits 0x%08X", mantissa);
//  "%#010x, ", mantissa);

if(exponent == 0)
{
    if(mantissa != 0)
    {
        printf("\ndenormalized  ");
    }
}
else{
    printf("\nnormalized:   ");
    tru_exp = exponent - 127;
    printf("exp = %d", tru_exp);
}

if(exponent == 0 && mantissa == 0 && negative == 1)
{

    printf("\n-zero");

}

if(exponent ==0 && mantissa == 0 && negative == 0)
{
 printf("\n+zero");
}



if(exponent == 255 && mantissa != 0 && negative == 1)
{

    printf("\nQNaN");

}

   if(exponent == 255 && mantissa != 0 && negative == 0)
{

    printf("\nSNaN");

}

if(exponent == 0xff && mantissa == 0 && negative == 1)
{
    printf("\n-infinity");
}

if(exponent == 0xff && mantissa == 0 && negative == 0)
{
    printf("\n+infinity");
}


    printf("\n");
}while(HexNumber != 0);



return 0;
  }

I dont think the de normalized is right?

Answer 1

Generally, you're pretty close. Some comments:

• 0x7fffffff is a quiet NaN, not a signaling NaN. The signbit does not determine whether or not a NaN is quiet; rather it is the leading bit of the significand (the preferred term for what you call "mantissa") field. 0xffbfffff is a signaling NaN, for example.

Edit: interjay correctly points out that this encoding isn't actually required by IEEE-754; a platform is free to use a different encoding for differentiating quiet and signaling NaNs. However, it is recommended by the standard:

A quiet NaN bit string should be encoded with the first bit of the trailing significand field T being 1. A signaling NaN bit string should be encoded with the first bit of the trailing significand field being 0.

• Infinities and NaNs usually aren't called "normal numbers" in the IEEE-754 terminology.

• Your condition for calling a number "denormal" is correct.

• For normal numbers, it would be nice to add the implicit leading bit when you report the significand. I personally would probably print them out in the C99 hex notation: 0x40000000 has a significand (once you add the implicit bit) of 0x800000 and an exponent of 1, so becomes 0x1.000000p1.

• I'm sure some aging PDP-11 hacker will give you a hard time about "big endian" and "little endian" not being the only two possibilities.

Edit Ok, example of checking for qNaN on platforms that use IEEE-754's recommended encoding:

if (exponent == 0xff && mantissa & 0x00400000) printf("\nqNaN");