Expression Evaluation Orders

This article will explain expression evaluation orders in all kinds of scenarios.

An Expression Is Evaluated After the Expressions It Depends On

This is easy to comprehend. An apparent example is an expression is evaluated later than its sub-expressions. For example, in a function call f(x, y[n]),

Please read program resource initialization order for another example on package-level variable initialization orders.

The Usual Order

For the evaluations within a function body, Go specification says
..., when evaluating the operands of an expression, assignment, or return statement, all function calls, method calls, and (channel) communication operations are evaluated in lexical left-to-right order.

The just described order is called the usual order.

Please note that an explicit value conversion T(v) is not a function call.

For example, in an expression []int{x, fa(), fb(), y}, assume x and y are two variables, fa and fb are two functions, then the call fa() is guaranteed to be evaluated (executed) before fb(). However, the following the evaluation orders are unspecified in Go specification: Another example, the following assignment, is demoed in Go specification.
y[z.f()], ok = g(h(a, b), i()+x[j()], <-c), k()
where Also considering the rule mentioned in the last section, compilers should guarantee the following evaluation orders at run time. However, the following orders (and more others) are not specified. By the usual order, we know the following declared variables x, m and n (also demoed in Go specification) will be initialized with ambiguous values.
	a := 1
	f := func() int { a++; return a }

	// x may be [1, 2] or [2, 2]: evaluation order
	// between a and f() is not specified.
	x := []int{a, f()}

	// m may be {2: 1} or {2: 2}: evaluation order between
	// the two map element assignments is not specified.
	m := map[int]int{a: 1, a: 2}

	// n may be {2: 3} or {3: 3}: evaluation order
	// between the key and the value is not specified.
	n := map[int]int{a: f()}

Evaluation and Assignment Orders in Assignment Statements

Beside the above introduced rules, Go specification specifies more on the expression evaluation order the order of individual assignments in an assignment statement:
The assignment proceeds in two phases. First, the operands of index expressions and pointer indirection (including implicit pointer indirection in selectors) on the left and the expressions on the right are all evaluated in the usual order. Second, the assignments are carried out in left-to-right order.

Later, we call the first phase as evaluation phase and the second phase as carry-out phase.

Go specification doesn't specify clearly how the operands in the mentioned expressions on the left should be exactly evaluated, which ever caused some disputes. So, here, this article will explain more on the evaluation orders in value assignments.

To make the following explanations convenient, we assume that the map value of a map index destination expression in an assignment is always addressable. If it is not, we can think the map value has already been saved in and replaced by a temporary addressable map value before executing the assignment.

At the time of the end of the evaluation phase and just before the carry-out phase starts, each destination expression on the left of an assignment has been evaluated as its elementary form. Different destination expressions have different elementary forms. Assume a and b are two addressable variables of the same type, the following assignment
	a, b = b, a
will be executed like the following steps:
// The evaluation phase:
P0 := &a; P1 := &b
R0 := a; R1 := b

// The elementary form: *P0, *P1 = R0, R1

// The carry-out phase:
*P0 = R0
*P1 = R1

Here is another example, in which x[0] instead of x[1] is modified.
	x := []int{0, 0}
	i := 0
	i, x[i] = 1, 2
	fmt.Println(x) // [2 0]
The decomposed execution steps for the line 3 shown below are like:
// The evaluation phase:
P0 := &i; P1 := &x[0];
R0 := 1; R1 := 2

// The elementary form: *P0, *P1 = R0, R1

// The carry-out phase:
*P0 = R0
*P1 = R1

An example which is a little more complex.
package main

import "fmt"

func main() {
	m := map[string]int{"Go": 0}
	s := []int{1, 1, 1}; olds := s
	n := 2
	p := &n
	s, m["Go"], *p, s[n] = []int{2, 2, 2}, s[1], m["Go"], 5
	fmt.Println(m, s, n) // map[Go:1] [2 2 2] 0
	fmt.Println(olds)    // [1 1 5]

The decomposed execution steps for the line 10 shown below are like:
// The evaluation phase:
P0 := &s; PM1 := &m; K1 := "Go"; P2 := p; P3 := &s[2]
R0 := []int{2, 2, 2}; R1 := s[1]; R2 := m["Go"]; R3 := 5
// now, R1 == 1, R2 == 0

// The elementary form: *P0, (*PM1)[K1], *P2, *P3 = R0, R1, R2, R3

// The carry-out phase:
*P0 = R0
(*PM1)[K1] = R1
*P2 = R2
*P3 = R3

The following example rotates all elements in a slice for one index.
	x := []int{2, 3, 5, 7, 11}
	t := x[0]
	var i int
	for i, x[i] = range x {}
	x[i] = t
	fmt.Println(x) // [3 5 7 11 2]

Another example:
	x := []int{123}
	x, x[0] = nil, 456        // will not panic
	x, x[0] = []int{123}, 789 // will panic

Although it is legal, it is not recommended to use complex multi-value assignments in Go, for their readabilities are not good and they have negative effects on both compilation speed and execution performance.

As above has mentioned, not all orders are specified in Go specification for value assignments, so some bad written code may produce different results. In the following example, the expression order of x+1 and f(&x) is not specified. So the example may print 100 99 or 1 99.
package main

import "fmt"

func main() {
	f := func (p *int) int {
		*p = 99
		return *p

	x := 0
	y, z := x+1, f(&x)
	fmt.Println(y, z)

The following is another example which will print ambiguous results. It may print 1 7 2, 1 8 2 or 1 9 2, depending on different compiler implementations.
package main

import "fmt"

func main() {
	x, y := 0, 7
	f := func() int {
		return x
	fmt.Println(f(), y, f())

Expression Evaluation Orders in switch-case Code Blocks

The expression evaluation order in a switch-case code block has been described before. Here just shows an example. Simply speaking, before a branch code block is entered, the case expressions will be evaluated and compared with the switch expression one by one, until a comparison results true.
package main

import "fmt"

func main() {
	f := func(n int) int {
		fmt.Printf("f(%v) is called.\n", n)
		return n

	switch x := f(3); x + f(4) {
	case f(5):
	case f(6), f(7), f(8):
	case f(9), f(10):

At run time, the f() calls will be evaluated by the order from top to bottom and from left to right, until a comparison results true. So f(8), f(9) and f(10) will be not evaluated in this example.

The output:
f(3) is called.
f(4) is called.
f(5) is called.
f(6) is called.
f(7) is called.

Expression Evaluation Orders in select-case Code Blocks

When executing a select-case code block, before entering a branch code block, all the channel operands of receive operations and the operands of send statements involved in the select-case code block are evaluated exactly once, in source order (from top to bottom, from left to right).

Note, the target expression being assigned to by a receive case operation will only be evaluated if that receive operation is selected later.

In the following example, the expression *fptr("aaa") will never get evaluated, for its corresponding receive operation fchan("bbb", nil) will not be selected.
package main

import "fmt"

func main() {
	c := make(chan int, 1)
	c <- 0
	fchan := func(info string, c chan int) chan int {
		return c
	fptr := func(info string) *int {
		return new(int)

	select {
	case *fptr("aaa") = <-fchan("bbb", nil): // blocking
	case *fptr("ccc") = <-fchan("ddd", c):   // non-blocking
	case fchan("eee", nil) <- *fptr("fff"):  // blocking
	case fchan("ggg", nil) <- *fptr("hhh"):  // blocking
The output of the above program:

Note that the expression *fptr("ccc") is the last evaluated expression in the above example. It is evaluated after its corresponding receive operation <-fchan("ddd", c) is selected.

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