哪种位模式表示双精度浮点的最大值？

Question

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?

Answer 1

指数是308，因为那是1.7976931x 10 ^ 308，也就是1x 2 ^ 1024。

Answer 2

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.

Answer 3

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

Ie 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 .