Why is “int i = 2147483647 + 1;” OK, but “byte b = 127 + 1;” is not compilable?

Question

Why is int i = 2147483647 + 1; OK, but byte b = 127 + 1; is not compilable?

Answer 1

JLS3 #5.2 Assignment Conversion

( variable = expression )

In addition, if the expression is a constant expression (§15.28) of type byte, short, char or int :

A narrowing primitive conversion may be used if the type of the variable is byte, short, or char, and the value of the constant expression is representable in the type of the variable.

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Without this clause, we wouldn't be able to write

byte x = 0;
char c = 0;

But should we be able to do this? I don't think so. There are quite some magic going on in conversion among primitives, one must be very careful. I would go out of my way to write

byte x = (byte)0;