patterngoCritical
Change values while iterating
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iteratingchangewhilevalues
Problem
Let's suppose I have these types:
and that I want to iterate on my node's attributes to change them.
I would have loved to be able to do:
but as
Is there a simpler or faster way? Is it possible to directly get pointers from
Obviously I don't want to change the structures just for the iteration and more verbose solutions are no solutions.
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.
Solution
No, the abbreviation you want is not possible.
The reason for this is that
The specification about range says:
So, range uses
value is copied, making the original value untouchable.
This behavior is demonstrated by the following code:
The code prints you completely different memory locations for the value from range and the actual
value in the slice:
So the only thing you can do is to either use pointers or the index, as already proposed by jnml and peterSO.
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] ESo, range uses
a[i] as its second value for arrays/slices, which effectively means that thevalue 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. 0x7f095ed0bf68So the only thing you can do is to either use pointers or the index, as already proposed by jnml and peterSO.
Code Snippets
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] Ex := make([]int, 3)
x[0], x[1], x[2] = 1, 2, 3
for i, val := range x {
println(&x[i], "vs.", &val)
}0xf84000f010 vs. 0x7f095ed0bf68
0xf84000f014 vs. 0x7f095ed0bf68
0xf84000f018 vs. 0x7f095ed0bf68Context
Stack Overflow Q#15945030, score: 219
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