max floating point value [duplicate]

Question

I am wondering if the max float represented in IEEE 754 is:

(1.11111111111111111111111)_b*2^[(11111111)_b-127]

Here _b means binary representation. But that value is 3.403201383*10^38, which is different from 3.402823669*10^38, which is (1.0)_b*2^[(11111111)_b-127] and given by for example c++ <limits>. Isn't (1.11111111111111111111111)_b*2^[(11111111)_b-127] representable and larger in the framework?

Does anybody know why?

Thank you.

Answer 1

The exponent 11111111_b is reserved for infinities and NaNs, so your number cannot be represented.

The greatest value that can be represented in single precision, approximately 3.4028235×10³⁸, is actually 1.11111111111111111111111_b×2^{11111110_b-127}.

See also http://en.wikipedia.org/wiki/Single-precision_floating-point_format

Answer 2

Being the "m" the mantissa and the "e" the exponent, the answer is:

In your case, if the number of bits on IEEE 754 are:

• 16 Bits you have 1 for the sign, 5 for the exponent and 10 for the mantissa. The largest number represented is 4,293,918,720.
• 32 Bits you have 1 for the sign, 8 for the exponent and 23 for the mantissa. The largest number represented is 3.402823466E38
• 64 Bits you have 1 for the sign, 11 for the exponent and 52 for the mantissa. The largest number represented is 2¹⁰²⁴ - 2⁹⁷¹