IEEE-754 Floating point --> Special values and Ranges

아티클
02/25/2006

As I mentioned in my previous blog, we shall now discuss details regarding range of values for IEEE-754 floating point numbers, Denormalized forms, NAN and a simple algorithms for conversion between IEEE-754 hexadecimal representations and decimal Floating-Point Numbers.

Let’s take a look at the special values in IEEE-754 floating point representation.

Special values:

Exponent field values of all 0s and all 1s are used to denote special values in this scheme

Zero

Zero is represented with an exponent field of zero and a mantissa field of zero. Depending on the sign bit, it can be a positive zero or a negative zero. Thus, -0 and +0 are distinct values, though they are treated as equal.

Infinity

Infinity is represented with an exponent of all 1s and a mantissa of all 0s. Depending on the sign bit, it can be a positive infinity(+¥) or negative infinity (-¥). The infinity is used in case of the saturation on maximum representable number so that the computation could continue.

NaN

The value NaN (Not a Number) is used to represent a value that does not represent a real number. They are used in computations that generate undefined results so that with NaN the operations are defined for it to let the computations continue. NaN's are represented by a bit pattern with an exponent of all 1s and a non-zero mantissa. There are two categories of NaN: QNaN (Quiet NaN) and SNaN (Signalling NaN).

A QNaN is a NaN with the most significant fraction bit set (denotes indeterminate operations).

An SNaN is a NaN with the most significant fraction bit clear (denotes invalid operations).

Denormalized

If the exponent is all 0s, and the mantissa is non-zero, then the value is treated as a denormalizednumber. The denormalized numbers does not have an assumed leading 1 before the binary point. For Single precision, this represents a number (-1)^s × 0.m × 2^-126, where s is the sign bit and m is the mantissa. For double precision, it represents as (-1)^s × 0.m × 2^-1022.

Thus, following are the values corresponding to a given representation:

(Note that b used in the table is the bias)

Sign(s)	Exponent (e)	Mantissa (m)	Range for Single Precision values in binary	Range Name
1	11..11	10..00:11.11	__	QNaN
1	11..11	00..01:01..11	__	SNaN
1	11..11	00..00	< -(2-2^-23) × 2¹²⁷	-Infinity (Negative Overflow)
1	11..10:00..01	11..11:00..00	-(2-2^-23) × 2¹²⁷:-2^-126	Negative Normalized -1.m × 2^(e-b)
1	00..00	11..11:00..01	-(1-2^-23) × 2^-126:-2^-149	Negative Denormalized -0.m × 2^(-b+1)
__	__	__	-2^-150:< -0	Negative Underflow
1	00..00	00..00	-0	-0
0	00..00	00..00	+0	+0
__	__	__	> +0:2^-150	Positive Underflow
0	00..00	00..01:11..11	2^-149: (1-2^-23) × 2^-126	Positive Denormalized 0.m × 2^(-b+1)
0	00..01:11..10	00..00:01..11	2^-126:(2-2^-23) × 2¹²⁷	Positive Normalized 1.m × 2^(e-b)
0	11..11	00..00	> (2-2^-23) × 2¹²⁷	+Infinity (Positvie Overflow)
0	11..11	00..01:01..11	__	SNaN
0	11..11	10..00:11.11	__	QNaN

Range:

As, mentioned in the table above, range for the positive normalized no for single precision float is

2^-126 to (2-2^-23) × 2¹²⁷ . Note that the bias(b) here is 127.

Let’s see how did we arrive we arrive at these ranges. As mentioned table above, the positive normalized form would be represented as 1.m × 2^(e-b) where m is mantissa, e is exponent and b is bias.

Thus, smallest normalized no for single precision would come out as 1.0…0( all 0’s after decimal) x 2^1-127 such that mantissa is 0 as and exponent is 1, thus:

1.0 x 2^1-127 à 2^-126

Now, the largest normalized no for single precision would come out as 1.1…..1( 23 1’s after decimal) x 2^254-127 such that mantissa is all ones and exponent is also all 1s except the least significant bit(254), thus this no equals :

1……1( 24 ones) 2²⁴ - 1

----------------------- x 2^254-127 à -------------- X 2¹²⁷ à (2-2^-23) × 2¹²⁷

2²³ 2²³

Again as mentioned in the table above, range for the positive denormalized no for single precision float is 2^-149 to (1-2^-23) × 2^-126 .Note that the bias(b) here is 127 and denormalized form would be represented as: 0.m × 2^(-b+1)

Thus, smallest denormalized no for single precision would come out as 0.00…..1 x 2^-127+1 such that mantissa has all the bits as 0 except the least significant bit and exponent is anyways 0, thus:

0.00…..1 x 2^-127+1 à 2^-23 x 2^-126 à 2^-149

And the largest denormalized no for single precision would come out as 0.11…..1 x 2^-127+1 such that mantissa has all the bits as 1 and exponent is anyways 0, thus:

1……1( 23 ones) 2²³ - 1

----------------------- x 2^-127+1 à -------------- X 2^-126 à (1-2^-23) × 2^-126

2²³ 2²³

Similarly, you can derive the ranges for double precision floats as well. The following table shows the ranges for the single as well the double precision floats for their positive as well as negative values.

	Single Precision	Double Precision
Normalized form	± 2^-126 to (2-2^-23)×2¹²⁷	± 2^-1022 to (2-2^-52)×2¹⁰²³
Denormalized form	± 2^-149 to (1-2^-23)×2^-126	± 2^-1074 to (1-2^-52)×2^-1022

Algorithms:

Let’s take a look into an algorithm(written in C++) which takes a 32-bit integer (which contains the simple bit representation for a single precision float) and returns an equivalent float value.

float single_float_from_storage_bits(int storagebits)

{

//Check the sign bit and assign the same to sign

int sign = ((storagebits & 0x80000000) == 0) ? 1 : -1;

// get the exponent value, bit postion 30 - 23

int exponent = ((storagebits & 0x7f800000) >> 23);

// get the mantissa value, bit position 22 - 00

int mantissa = (storagebits & 0x007fffff);

//if exponent is 0, it could be either 0 or denormalized form.

if (exponent == 0)

{

// since matissa is also 0, definitely this is a 0

if (mantissa == 0)

{

// We would decide +ve or -ve 0 depending on sign

return (sign * 0.0f);

}

else

// else return the calculated denormalized value

return (float)(sign * mantissa * pow(2, -149));

}

//if exponent is all 1, then it could be either Infinity or NaN

else if (exponent == 0xff)

{

//if mantissa is 0, then it is +infnity or -infinity

if (mantissa == 0)

{

// Use sign to decide +infnity or -infinity

return ((sign == 0) ? -INFINITY : +INFINITY) ;

}

// Else its a NaN, you can also check SNaN or QNan here

else return NaN;

}

// Now we are sure this is a Normalized no

else

{

// add the implied 24th bit of mantissa here

mantissa |= 0x00800000;

// return the normalized form here

return (float)(sign * mantissa * pow(2, exponent-150));

}

In a similar manner, you can do the vice versa, i.e given a single precision float, you can find out its storage bits representation layout.

Comments

Anonymous
April 24, 2006
I found these blog entries from Prem to be quite exhaustive. So posting the links here just in case you...
Anonymous
May 05, 2007
Maybe someone could give me (a C novice, with little exposure to IEEE Floating-Point Numbers) some feedback on a solution I have used to the converting a 32 bit integer to a floating point value. Please note that my application is based on the use of Z-World’s Dynamic C and I have no idea whether it might be successful on other platforms The basis of the conversion is that the unsigned long 32 bit integer (k) formats to the equivalent floating point value if it is output using the %f conversion character. unsigned long k; float val; char valstr[32]; void main() { k = // IEEE floating point number sprintf(valstr,"%f", k); val=atof(valstr); // val returns equivalent float value for k }
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May 27, 2007
The comment has been removed
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Anonymous
July 27, 2015
Thanks a lot....Very very helpful to me..
Anonymous
September 09, 2015
Why do you use 150 as your bias and not 127? S/b [1 << [exp bit count - 1]] - 1.

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IEEE-754 Floating point --> Special values and Ranges

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