Go中的同时变量分配不同于单个变量分配

Question

I was under the impression that despite the differences in syntax, function a and function b below were logically equivalent. However, they are not and I do not understand the difference between them.

It seems to me that they are both assigning:

• the value of x to the variable z,
• the value of y to the variable x, and
• the value of x+y to the variable y.

Could anyone help clear up my misunderstanding regarding the multiple variable assignment and the logical difference between function a and function b?

package main

import "fmt"

func a() (int, int, int) {
    x:=1
    y:=2
    z:=3

    z = x
    x = y
    y = x+y

    return x, y, z
}

func b() (int, int, int) {
    x:=1
    y:=2
    z:=3

    z, x, y = x, y, x+y

    return x, y, z
}

func main() {
    fmt.Println(a()) // prints 2 4 1
    fmt.Println(b()) // prints 2 3 1
}

Answer 1

Assignment can be thought of as an "atomic" operation. That is, it's useful to think that all values on the left hand side of the = are "frozen" until all of the operations are finished.

Consider the following program:

package main

import "fmt"

func swap() (int, int) {
    x := 1
    y := 2
    x, y = y, x
    return x, y
}

func main() {
    fmt.Println(swap()) // prints 2 1
}

Without this "freezing" behaviour, you would get 2 for both x and y, which is probably not what you'd expect from the code. It's also probably easier to reason about the semantics of this "freezing" behaviour than if the "cascading" approach were taken.