Java implicit conversion

Question

With the following code:

Float a = 1.2;

there is an error because it takes the decimal as double value and double is a bigger datatype than float.

Now, it takes integer as default int type. So, why is the following code not giving any error?

Byte b = 20;

Answer 1

The compiler is smart enough to figure out that the bit representation of 20 (an int value) can fit into a byte with no loss of data. From the Java Language Specification §5.1.3:

A narrowing primitive conversion from double to float is governed by the IEEE 754 rounding rules (§4.2.4). This conversion can lose precision, but also lose range, resulting in a float zero from a nonzero double and a float infinity from a finite double. A double NaN is converted to a float NaN and a double infinity is converted to the same-signed float infinity.

A narrowing conversion of a signed integer to an integral type T simply discards all but the n lowest order bits, where n is the number of bits used to represent type T. In addition to a possible loss of information about the magnitude of the numeric value, this may cause the sign of the resulting value to differ from the sign of the input value.

See also this thread.