CODEWITHSUNDEEP
go
Harry
Harry

Reputation: 146

swap function not working in golang

Actually i just start to learn golang . In the beginning i think that = and := are same . But then i understand that there is some difference between this two .

I learned swap function in golnag 

import "fmt"

func swap(x, y string) (string, string) {
    return y, x
}

func main() {
    a, b := swap("hello", "world")
    fmt.Println(a, b)
}

But when i rewrite this function using var this is not working 

package main

import "fmt"

func swap(x, y string) (string, string) {
    return y, x
}

func main() {
   var a, b string
    a ="hello"
    b="world"
     swap(a, b)
    fmt.Println(a, b)
}

what is the error in this program ?

Upvotes: 0

Views: 3789

Answers (4)

I159
I159

Reputation: 31139

Variables declaration

  • var a string - declaration of a variable with null value
  • a := "spam" - declaration of a variable with a concrete value
  • func f(a, b string) (string, string) { - declaration of a function with value parameters. It means you have new variables with passed values as arguments each time you call a function.
  • func f(a, b *string) (*string, *string) { - declaration of a function with pointer arguments. In it's turn it means you have pointers to passed variables each time you call the function.

Also...

  • a := *string - declaration of a pointer variable.
  • *a - value of a pointer variable.
  • &a - pointer of a value

In-place swap

To swap in-place (without returning and reassigning) you should swap values between pointers.

func swap(a, b *string) {
    *a, *b = *b, *a
}

p.s.

Take into account that strings is read-only slices of bytes. And slices are reference type it means that an array behind the sub-slices of a common array or slice is the same. It doesn't related to the question but should be considered in such cases.

Upvotes: 2

md2perpe
md2perpe

Reputation: 3071

Another solution is to use pointers:

package main

import "fmt"

func swap(x, y *string) {
    *x, *y = *y, *x
}

func main() {
    var a, b string
    a ="hello"
    b="world"
    swap(&a, &b)
    fmt.Println(a, b)
}

https://play.golang.org/p/-vxUMlaVmN

Upvotes: 8

omu_negru
omu_negru

Reputation: 4770

To respond your initial question, you should assign the values returned by swap to a and b like so

a, b = swap(b, a)

Notice that this is simple assignment , without the : attached to the equal

also, instead of a swap function, you could just try inplace reassignment: a, b = b, a

Upvotes: 3

Eugene Lisitsky
Eugene Lisitsky

Reputation: 12845

The reason is that in second case values returned from swap are ignored. SO nothing is changed.

Try: https://play.golang.org/p/uADEf5X15g

package main

import "fmt"

func swap(x, y string) (string, string) {
    return y, x
}

func main() {
    var a, b string
    a = "hello"
    b = "world"
    a, b = swap(a, b)  ////   <----
    fmt.Println(a, b)
}

Upvotes: 2

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