Why double.MaxValue is larger than long.MaxValue?

问题

both of them hold 8 bytes, but how come the max value for double is much greater than the max value of long? there is a finite number of bits available, so how could you reach greater numbers with floating point variables?

回答1:

It uses a different representation (floating point) using exponents and mantissa

For details see IEEE754

回答2:

A double has something called an exponent, which is basically just a scaling factor. This allows the range of double to be much greater, but at the cost of precision.

A long is a simple integer value with no scaling factor.

回答3:

Because floating-point representation is of lower precision. While long type can represent all integer numbers in the range from its minimum to maximum, double type can only represent some of it.

Since they occupy same amount of bits, the amount of numbers each is capable to express are nearly equal (actually, double can represent fewer numbers). Just the paces between these numbers are different.

回答4:

Floating point numbers consist of a mantissa and an exponent, and the value of a floating point number is:

mantissa * 2^exponent

The exponent in a Double is 11 bits, so the maximum value is of the magnitude 2^{211 - 1} = 2¹⁰²⁴, which is way more than the magnitude of a 64-bit signed double, which is 2⁶³-1.

来源：https://stackoverflow.com/questions/2639941/why-double-maxvalue-is-larger-than-long-maxvalue