Byte.toBinaryString

Question
1 
 Hello,

isn’t a Byte supposed to consist of eight bits?

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scala> val b: Byte = -1
b: Byte = -1

scala> b.toBinaryString
res1: String = 11111111111111111111111111111111
1 Like
2

In explicit form, that’s intWrapper(b).toBinaryString - this is a method on RichInt, operating on Int.

3

Is there a particular reason why we don’t warp into a RichByte with a toBinaryString method in it?

1 Like
4

Likely a combination of pragmatic reasons, among them that, while the RichInt method simply delegates to Java’s java.lang.Integer, there’s no such method on java.lang.Byte.