spec: order of evaluation of variables in return statement is not determined

[zboya]

Please answer these questions before submitting your issue. Thanks!

What version of Go are you using (go version)?

go version
go version go1.10.2 darwin/amd64

What did you do?

package main

import "fmt"

func main() {
	m, i, j := testChange()
	fmt.Println("retun ", m, i, j)
}

func testChange() (int, interface{}, int) {
	m := 0
	i := 0

	f := func() int {
		m = 2
		i = 2
		return 99
	}

	return m, i, f()
}

What did you expect to see?

retun 2 2 99

What did you see instead?

retun 2 0 99

[cznic]

This seems to be working as specified.

edit: Oversight, no more I think the above. 0 0 99 or 2 2 99.

[DeedleFake]

How is this working as specified? This seems to show a discrepancy in the behavior of assigning to a variable depending on whether it's an interface or not. Not trying to be accusatory; just curious what part of the spec covers this.

Edit: Better wording.

[cznic]

Yup, I edited my post while you were writing yours. My apologies.

[fraenkel]

There is no guaranteed order, see https://golang.org/ref/spec#Order_of_evaluation

[cznic]

That seems to specify evaluating of a single expression. The order of evaluation of the expressions in the list of expressions of an return statement seems to be unspecified, so the evaluating m, i, f() in order i, f(), m would warrant the OP observed result.

I will now stop myself from polluting this issue. I'm no more sure about anything involved in it.

[ianlancetaylor]

With Go 1.3 and later I get 2 0 99. With gccgo I get 0 0 99. With Go 1.0 to 1.3 I get 2 2 99.

I think this boils down to when we evaluate variables listed in the return statement. Do we load m and i before or after the call to f? I don't think the language spec currently specifies that. In other words, I think this is another example of #23188 (comment) .

I don't think the detail of the fact that this is an interface value affects this. It happens to change what the compiler does, but I don't think any specific ordering is required anyhow.

[KISSMonX]

go1.10.2 darwin/amd64

[KISSMonX]

Test Code: https://play.golang.org/p/xnHqSQyrp_J

package main

import "fmt"

func main() {
	m, i, j := testChange00()
	fmt.Println("retun ", m.(int), i.(int), j.(int)) // 0 0 99
	fmt.Println("retun ", m, i, j)                   // 0 0 99

	a, b, c := testChange11()
	fmt.Println("retun ", a, b, c) // 2 2 99
}

// return type: int int int  				Result: 2 2 99
// return type: int int interface{} 			Result: 2 2 99
// return type: int interface{}, interface{} 		Result: 2 0 99
// return type: int interface{}, int 			Result: 2 0 99
// return type: interface{} interface{} interface{} 	Result: 0 0 99
// return type: interface{} interface{} int 		Result: 0 0 99
// return type: interface{} int int 			Result: 0 2 99
// return type: interface{} int interface{}		Result: 0 2 99
func testChange00() (interface{}, interface{}, interface{}) {
	m := 0
	i := 0

	f := func() int {
		m = 2
		i = 2
		return 99
	}

	return m, i, f()
}

func testChange11() (int, int, int) {
	m := 0
	i := 0

	f := func() int {
		m = 2
		i = 2
		return 99
	}

	return m, i, f()
}

[zboya]

@KISSMonX that is the reason why I wonder, I think the interface dose affect the result.

[KISSMonX]

Yeah, indeed affected, but return type should not affect the order of evaluation.
Waiting for official reply. 😃

[ianlancetaylor]

With the current spec, the order of evaluation is unpredictable. There is no one correct answer, there are several correct answers. Correct Go implementations can produce different results for this program. As I said above, different compilers behave differently today. It may happen that with the current Go 1.10 compiler the order is affected by whether you use an interface type, but future Go releases may well produce different results that are still completely valid.