operator 3

package main

const N = 1
var n = N
var a byte = 128 << N >> N
var b byte = 128 << n >> n

func main() {
	println(a, b)
}

Choices:

• 128 128
• 128 0
• 0 128
• 0 0

Answer: 128 0

Run it on Go play.

Key points:

• If the operands in an operator expression are both/all constants, then the expression is evaluated at compile time. In the above program, 128 << N >> N is such an expression. In this expression, 128 is deduced as an untyped int value.
• In a bit-shift expression, if the left operand is an untyped constant and the right operand is not constant, then the type of the left operand will be deduced as the final assumed type. In the above program, 128 << n >> n is a such expression. In this expression, the type of 128 is deduced as the assumed type, byte. 128 << n overflows the value range of byte, so it is truncated to 0.

This is one small detail in Go programming. Have interests on more ones? Please read Go Details & Tips 101 book.

• slice: 1, 2, 3
• map: 1
• channel: 1
• loop: 1, 2, 3, 4
• switch: 1
• defer: 1, 2
• panic/recover: 1, 2
• reflect: 1, 2
• const: 1, 2, 3
• operator: 1, 2, 3
• function call: 1
• scope: 1, 2
• nil: 1
• embedding: 1

(more are coming ...)