slice 2

package main

func main() {
	var x = []string{"A", "B", "C"}

	for i, s := range x {
		print(i, s, ",")
		x[i+1] = "M"
		x = append(x, "Z")
		x[i+1] = "Z"
	}
}

Choices:

• 0A,1B,2C,
• 0A,1Z,2Z,
• 0A,1M,2M,
• 0A,1M,2C,
• 0A,1Z,2M,
• 0A,1M,2Z,
• (infinite loop)

Answer: 0A,1M,2C,

Run it on Go play.

Key points:

• Ranging over a container iterates the elements of a copy of the container. The evaluation of the copy happens before executing the loop, so the length and capacity of the copy are never changed.
• A slice references its elements on a backing array. So a copy of a slice shares the same elements (and the backing array) with the slice.
• The assignment x[i+1] = "M" in the first loop step modifies the second element of the initial slice x and the copy of the initial slice x.
• If the free element slots of the first argument slice of an append call are insufficient to hold all the appended elements, the append call will create a new backing array to hold all the elements of the first argument slice and the appended elements. So, at the end of the first loop step, the backing array of the slice x is changed. However, the change doesn't affect the slice copy used in the element iteration. All subsequent element modifications apply to the new backing array, so they have no effects on the copy used in the element iteration.

This above quiz is extended from one of Valentin Deleplace's quizzes. The following is the original quiz.

package main

func main() {
	var x = []string{"A", "B", "C"}

	for i, s := range x {
		print(i, s, " ")
		x = append(x, "Z")
		x[i+1] = "Z"
	}
}

The original quiz prints 0A,1B,2C,, because the ranged container is a copy of the initial x, and elements of the copy are never changed.

The following is another quiz made by Valentin Deleplace (with a bit modification).

package main

func main() {
	var y = []string{"A", "B", "C", "D"}
	var x = y[:3]

	for i, s := range x {
		print(i, s, ",")
		x = append(x, "Z")
		x[i+1] = "Z"
	}
}

The other quiz prints 0A,1Z,2C,. It is similar to the above extended quiz. Key points:

• The first append call doesn't create a new backing array, so the assignment x[i+1] = "Z" in the first loop has effect on the iniital slice x (and its copy used in the element iteration).
• The second append call creates a new backing array, so subsequent x[i+1] = "Z" assignments have no effects on the iniital slice x (and its copy used in the element iteration).

• 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 ...)