How to calculate the range of data type float in c++?

Question

As we can see int has 4 byte in memory, that are 32bits, after applying range formula , we can see range of int -2147483648 to 2147483647. I have calculated the ranges of all datatypes besides float and double and long double. I dont know how they calculated the range of float mentioned below.

Answer 1

Floating point numbers are stored as an exponent and a fraction within the space available.

For some systems where float is implemented as an IEEE 754 value, the results would looks as below.

sign : 1 bit
exponent : 8 bits
fraction : 23 bits

The exponent allows numbers from 2 ^ (-127) (2 to the power -127) to 2 ^ 128 ( 2 to the power 128).

Allowing a range of numbers from

5.87747E-39 3.40282E+38

the fraction point gives a fraction such as .12313

Thus with 23 bits of values, the accuracy of a number is about 7 decimal digits or 1.19 E-7

For more details see wikipedia : IEEE 754-1985

On a given system, the <cfloat> / <float.h> will give the limits. For non IEEE 754 based representations, you would have to understand how the numbers are stored to calculate the limits.

Answer 2

-2^(n-1) to (2^(n-1)-1) is the formula to calculate the range of data types.

Where n = no.of.bits of the primitive data type.

For example: for the byte data type, n = 8 bits

-2^(8-1) to (2^(8-1)-1)

The above calculation will give you -128 to 127. Now, coming to the question of why it’s not 255. The reason is that byte, int, short, and double are signed data types meaning it has half the range below 0 (negative) and half the range above 0 (positive). The first bit represents a sign (+ or -). The remaining bits are 7. That’s why 2^(8-1) = 128. We take 0 as a positive sign, so the range is 2^(8-1) - 1 for positive numbers.