How is the range of int values in Java calculated?

The first definition of an int in the JDK is 4 bytes ===> 32 bits (all subsequent calculations are based on this)

The 32-bit is the JVM that allocates only 32 squares of space for storing data.

It is well known that computers store data in 0 and 1.

Then, the 32-lattice 0 or 1 method has 2 32-time species:

So. In these 32 squares. Or a 32-bit space that identifies 10 decimal digits:

Minimum

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

Biggest

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

The first method of calculation: The binary maximum number (32 1) is converted to 10 binary, that is, 4294967296;

Another method of calculation: Since there are 2 of the 32-way algorithm, then the largest number according to the 10 binary is 2 32 times. i.e. 4294967296;

Pity. The above calculation is unsigned. is a positive number. But in Java, there are positive and negative points in Int. So 32 squares occupy a lattice mark plus or minus.

So only 31 squares can be used to identify values.

：

X

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

X is 0 to identify positive or negative.

The last int can be identified by the maximum/minimum number is: 2 of 31 Square: +/-2147483648

How is the range of int values in Java calculated?