Reputation: 105517

What bit pattern represents the max value for double precision floating point?

I'm wondering how the max value is represented in 64 bits of double precision floating point. I assume it's represented with all 1's in exponent and mantissa like so:

0 11111111111 1111111111111111111111111111111111111111111111111111

If so, then why Number.MAX_VALUE = 1.7976931348623157e+308 shows the exponent of 308, instead of 1024 decoded from 11111111111? Is the bits pattern different?

Upvotes: 1

Answers (3)

Zbynek Vyskovsky - kvr000

Reputation: 18835

308 is decimal exponent while double uses powers of two.

Secondly max value exponent is reserved for infinity and NaN. Max-1 is the max exponent for regular number. Therefore the max number is written as:

01111111 11101111 11111111 11111111 11111111 11111111 11111111 11111111

I.e. 0 sign bit, 2046 as exponent value (note that 1023 means actually zero exponent so 2046 means exponent of 1023) and mantissa all ones (1.11111(52 times) in binary, the first one is hidden), in other words 1.11111(52 times)*2^1023.

Converted to decimal it's (2-(2^-52))*2^1023 which is about 1.79769313486231*10^308.

For the double/float formats you can find very precise information on wiki.

Upvotes: 3

user6551188

Reputation:

The exponent is 308 since 308 is 1024 / (log 10 / log 2).

Remember that the 1024 is a binary exponent, but 308 is a denary exponent.

Upvotes: 2