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:
Answer: 0A,1M,2C,
Run it on Go play.
Key points:
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.
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:
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).
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).
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