Not getting expected results with waitgroup [duplicate] - go

Question

Let's suppose I have these types:
type Attribute struct {
Key, Val string
}
type Node struct {
Attr []Attribute
}
and that I want to iterate on my node's attributes to change them.
I would have loved to be able to do:
for _, attr := range n.Attr {
if attr.Key == "href" {
attr.Val = "something"
}
}
but as attr isn't a pointer, this wouldn't work and I have to do:
for i, attr := range n.Attr {
if attr.Key == "href" {
n.Attr[i].Val = "something"
}
}
Is there a simpler or faster way? Is it possible to directly get pointers from range?
Obviously I don't want to change the structures just for the iteration and more verbose solutions are no solutions.

Answer 1

No, the abbreviation you want is not possible.
The reason for this is that range copies the values from the slice you're iterating over.
The specification about range says:
Range expression 1st value 2nd value (if 2nd variable is present)
array or slice a [n]E, *[n]E, or []E index i int a[i] E
So, range uses a[i] as its second value for arrays/slices, which effectively means that the
value is copied, making the original value untouchable.
This behavior is demonstrated by the following code:
x := make([]int, 3)
x[0], x[1], x[2] = 1, 2, 3
for i, val := range x {
println(&x[i], "vs.", &val)
}
The code prints you completely different memory locations for the value from range and the actual
value in the slice:
0xf84000f010 vs. 0x7f095ed0bf68
0xf84000f014 vs. 0x7f095ed0bf68
0xf84000f018 vs. 0x7f095ed0bf68
So the only thing you can do is to either use pointers or the index, as already proposed by jnml and peterSO.

Answer 2

You seem to be asking for something equivalent to this:
package main
import "fmt"
type Attribute struct {
Key, Val string
}
type Node struct {
Attr []Attribute
}
func main() {
n := Node{
[]Attribute{
{"key", "value"},
{"href", "http://www.google.com"},
},
}
fmt.Println(n)
for i := 0; i < len(n.Attr); i++ {
attr := &n.Attr[i]
if attr.Key == "href" {
attr.Val = "something"
}
}
fmt.Println(n)
}
Output:
{[{key value} {href http://www.google.com}]}
{[{key value} {href something}]}
This avoids creating a--possibly large--copy of type Attribute values, at the expense of slice bounds checks. In your example, type Attribute is relatively small, two string slice references: 2 * 3 * 8 = 48 bytes on a 64-bit architecture machine.
You could also simply write:
for i := 0; i < len(n.Attr); i++ {
if n.Attr[i].Key == "href" {
n.Attr[i].Val = "something"
}
}
But, the way to get an equivalent result with a range clause, which creates a copy but minimizes slice bounds checks, is:
for i, attr := range n.Attr {
if attr.Key == "href" {
n.Attr[i].Val = "something"
}
}

Answer 3

I'd adapt your last suggestion and use the index-only version of range.
for i := range n.Attr {
if n.Attr[i].Key == "href" {
n.Attr[i].Val = "something"
}
}
It seems simpler to me to refer to n.Attr[i] explicitly in both the line that tests Key and the line that sets Val, rather than using attr for one and n.Attr[i] for the other.

Answer 4

For example:
package main
import "fmt"
type Attribute struct {
Key, Val string
}
type Node struct {
Attr []*Attribute
}
func main() {
n := Node{[]*Attribute{
&Attribute{"foo", ""},
&Attribute{"href", ""},
&Attribute{"bar", ""},
}}
for _, attr := range n.Attr {
if attr.Key == "href" {
attr.Val = "something"
}
}
for _, v := range n.Attr {
fmt.Printf("%#v\n", *v)
}
}
Playground
Output
main.Attribute{Key:"foo", Val:""}
main.Attribute{Key:"href", Val:"something"}
main.Attribute{Key:"bar", Val:""}
Alternative approach:
package main
import "fmt"
type Attribute struct {
Key, Val string
}
type Node struct {
Attr []Attribute
}
func main() {
n := Node{[]Attribute{
{"foo", ""},
{"href", ""},
{"bar", ""},
}}
for i := range n.Attr {
attr := &n.Attr[i]
if attr.Key == "href" {
attr.Val = "something"
}
}
for _, v := range n.Attr {
fmt.Printf("%#v\n", v)
}
}
Playground
Output:
main.Attribute{Key:"foo", Val:""}
main.Attribute{Key:"href", Val:"something"}
main.Attribute{Key:"bar", Val:""}