Java's Primitive Data Types

boolean: 1-bit. May take on the values true and false only.

byte: 1 signed byte (two's complement). Covers values from -128 to 127.

short: 2 bytes, signed (two's complement), -32,768 to 32,767

int: 4 bytes, signed (two's complement). -2,147,483,648 to 2,147,483,647.

long: 8 bytes signed (two's complement). Ranges from -9,223,372,036,854,775,808 to +9,223,372,036,854,775,807.

float: 4 bytes, IEEE 754. Covers a range from 1.40129846432481707e-45 to 3.40282346638528860e+38 (positive or negative).

double: 8 bytes IEEE 754. Covers a range from 4.94065645841246544e-324d to 1.79769313486231570e+308d (positive or negative).

char: 2 bytes, unsigned, Unicode, 0 to 65,535

Java uses the IEEE-754 binary64 format. In this format, one bit represents a sign (+ or − according to whether the bit is 0 or 1), eight bits are used for an exponent, and 52 bits are used for the primary encoding of the significant. One bit of the 53-bit significand is encoded via the exponent.

The values of the exponent field range from 0 to 2047. 2047 is reserved for use with infinities and NaNs. 0 is used for subnormal numbers. The values 1 to 2046 are used for normal numbers. In this range, an exponent field value of E represents an exponent e = E−1023. So the lowest value of e is 1−1023 = −1022, and the highest value is 2046−1023 = 1023. An exponent field value of 0 also represents the lowest value of E, −1022.

The 53-bit significand represents a binary numeral d.ddd…ddd2, where the first d is 0 if the exponent field is 0 and 1 if the exponent field is 1 to 2046. There are 52 bits after the “.”, and they are given by the primary significand field.

Let S be the sign bit field, E be the value of the exponent field, and F be the value of the primary significand field as an integer. Then the value represented is:

• if E is 1 to 2046, (−1)^S • 2^E−1023 • (1 + F•2⁻⁵²),
• if E is 0, (−1)^S • 2⁻¹⁰²² • (0 + F•2⁻⁵²).

Now we can see the smallest positive number is represented when S is 1, E is 0, and F is 1 (00000000000000000000000000000000000000000000000000012). Then the value represented is (−1)⁰ • 2⁻¹⁰²² • (0 + 1•2⁻⁵²) = +1 • 2⁻¹⁰²² • 2⁻⁵² = 2⁻¹⁰⁷⁴.

The greatest finite number is represented when S is 1, E is 2046, and F is 2⁵²−1 (11111111111111111111111111111111111111111111111111112). Then the value represented is (−1)⁰ • 2^2046−1023 • (1 + (2⁵²−1)•2⁻⁵² = +1 • 2¹⁰²³ • (1 + 1 − 2⁻⁵²) = 2¹⁰²³ • (2¹ − 2⁻⁵²) = 2¹⁰²⁴ − 2⁹⁷¹.