Reassigning a slice parameter acts differently
package main
import "fmt"
func main() {
paths := string{"hello", "world", "mars"}
var result = delete(paths, 1)
fmt.Println(result)
fmt.Println(paths)
}
func delete(paths string, index int) string {
paths = append(paths[:index], paths[index+1:]...)
return paths
}
The result of the code above is the following:
[hello mars]
[hello mars mars]
As you see, the second fmt.Println(paths) obviously uses the modified slice but does not use the reassigned value. Why is that? I was expecting it to print [hello mars] like in the print before.
I know that paths being passed is not the same slice as paths parameter in the delete() function expect for referencing the same underlying array. But I still don't understand how I changed the underlying array of the paths being passed to delete function as it prints [hello mars mars] instead of [hello world mars].
go
asked Jan 1 at 1:57
TarikTarik
36.4k67205309
add a comment |
package main
import "fmt"
func main() {
paths := string{"hello", "world", "mars"}
var result = delete(paths, 1)
fmt.Println(result)
fmt.Println(paths)
}
func delete(paths string, index int) string {
paths = append(paths[:index], paths[index+1:]...)
return paths
}
The result of the code above is the following:
[hello mars]
[hello mars mars]
As you see, the second fmt.Println(paths) obviously uses the modified slice but does not use the reassigned value. Why is that? I was expecting it to print [hello mars] like in the print before.
I know that paths being passed is not the same slice as paths parameter in the delete() function expect for referencing the same underlying array. But I still don't understand how I changed the underlying array of the paths being passed to delete function as it prints [hello mars mars] instead of [hello world mars].
go
asked Jan 1 at 1:57
TarikTarik
36.4k67205309
add a comment |
package main
import "fmt"
func main() {
paths := string{"hello", "world", "mars"}
var result = delete(paths, 1)
fmt.Println(result)
fmt.Println(paths)
}
func delete(paths string, index int) string {
paths = append(paths[:index], paths[index+1:]...)
return paths
}
The result of the code above is the following:
[hello mars]
[hello mars mars]
As you see, the second fmt.Println(paths) obviously uses the modified slice but does not use the reassigned value. Why is that? I was expecting it to print [hello mars] like in the print before.
I know that paths being passed is not the same slice as paths parameter in the delete() function expect for referencing the same underlying array. But I still don't understand how I changed the underlying array of the paths being passed to delete function as it prints [hello mars mars] instead of [hello world mars].
go
asked Jan 1 at 1:57
TarikTarik
36.4k67205309
package main
import "fmt"
func main() {
paths := string{"hello", "world", "mars"}
var result = delete(paths, 1)
fmt.Println(result)
fmt.Println(paths)
}
func delete(paths string, index int) string {
paths = append(paths[:index], paths[index+1:]...)
return paths
}
The result of the code above is the following:
[hello mars]
[hello mars mars]
As you see, the second fmt.Println(paths) obviously uses the modified slice but does not use the reassigned value. Why is that? I was expecting it to print [hello mars] like in the print before.
I know that paths being passed is not the same slice as paths parameter in the delete() function expect for referencing the same underlying array. But I still don't understand how I changed the underlying array of the paths being passed to delete function as it prints [hello mars mars] instead of [hello world mars].
go
go
asked Jan 1 at 1:57
TarikTarik
36.4k67205309
asked Jan 1 at 1:57
TarikTarik
36.4k67205309
asked Jan 1 at 1:57
TarikTarik
36.4k67205309
asked Jan 1 at 1:57
TarikTarik
36.4k67205309
asked Jan 1 at 1:57
TarikTarik
36.4k67205309
36.4k67205309
add a comment |
add a comment |
1 Answer
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Because, as you said, the same underlying array is in use. When you do the append, paths[:1] is a slice of length 1 and capacity 3, and paths[2:] is a slice of length 1, so there is enough room in the underlying array of the first slice to append the new value without allocating a new array. paths in main is still a slice of length 3 since it was never modified, but since the underlying array was modified (specifically element 1), you see the value that you see.
You may want to take a look at https://blog.golang.org/go-slices-usage-and-internals if you haven't already.
answered Jan 1 at 2:22
Andy SchweigAndy Schweig
4,6602916
add a comment |
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1 Answer
1
active
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1 Answer
1
active
oldest
votes
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oldest
votes
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oldest
votes
Because, as you said, the same underlying array is in use. When you do the append, paths[:1] is a slice of length 1 and capacity 3, and paths[2:] is a slice of length 1, so there is enough room in the underlying array of the first slice to append the new value without allocating a new array. paths in main is still a slice of length 3 since it was never modified, but since the underlying array was modified (specifically element 1), you see the value that you see.
You may want to take a look at https://blog.golang.org/go-slices-usage-and-internals if you haven't already.
answered Jan 1 at 2:22
Andy SchweigAndy Schweig
4,6602916
add a comment |
Because, as you said, the same underlying array is in use. When you do the append, paths[:1] is a slice of length 1 and capacity 3, and paths[2:] is a slice of length 1, so there is enough room in the underlying array of the first slice to append the new value without allocating a new array. paths in main is still a slice of length 3 since it was never modified, but since the underlying array was modified (specifically element 1), you see the value that you see.
You may want to take a look at https://blog.golang.org/go-slices-usage-and-internals if you haven't already.
answered Jan 1 at 2:22
Andy SchweigAndy Schweig
4,6602916
add a comment |
Because, as you said, the same underlying array is in use. When you do the append, paths[:1] is a slice of length 1 and capacity 3, and paths[2:] is a slice of length 1, so there is enough room in the underlying array of the first slice to append the new value without allocating a new array. paths in main is still a slice of length 3 since it was never modified, but since the underlying array was modified (specifically element 1), you see the value that you see.
You may want to take a look at https://blog.golang.org/go-slices-usage-and-internals if you haven't already.
answered Jan 1 at 2:22
Andy SchweigAndy Schweig
4,6602916
Because, as you said, the same underlying array is in use. When you do the append, paths[:1] is a slice of length 1 and capacity 3, and paths[2:] is a slice of length 1, so there is enough room in the underlying array of the first slice to append the new value without allocating a new array. paths in main is still a slice of length 3 since it was never modified, but since the underlying array was modified (specifically element 1), you see the value that you see.
You may want to take a look at https://blog.golang.org/go-slices-usage-and-internals if you haven't already.
answered Jan 1 at 2:22
Andy SchweigAndy Schweig
4,6602916
answered Jan 1 at 2:22
Andy SchweigAndy Schweig
4,6602916
answered Jan 1 at 2:22
Andy SchweigAndy Schweig
4,6602916
answered Jan 1 at 2:22
Andy SchweigAndy Schweig
4,6602916
4,6602916
add a comment |
add a comment |
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