Smooth operator


package main

import "fmt"

func foo() (int, error) {
    return 10, nil

func main() {
    dict := make(map[string]int)
    dict["foo"] = 10

    // Valid
    fmt.Printf("Foo: %d\n", dict["foo"])
    // Also valid
    val, exists := dict["foo"]
    fmt.Printf("Foo: %d, exists: %v\n", val, exists)
    // Invalid
    fmt.Printf("Foo: %d\n", foo())
    // Valid
    val, err := foo()
    fmt.Printf("Foo: %d, error: %v\n", val, err)

Why does the map access operator get special treatment? It returns multiple values, but you're allowed to pretend it doesn't.

If you try the same thing with a function call, you get an error like multiple-value foo() (value of type (int, error)) in single-value context

One could imagine a user-defined map that has similar semantics, or something like a stack:

type Stack struct {
    store []interface{}

func (s *Stack) push(obj interface{}) {
    s.store = append(s.store, obj)

func (s *Stack) pop() (interface{}, bool) {
    if len(s.store) == 0 {
        return nil, false
    last := len(s.store) - 1
    ret := s.store[last]
    s.store = s.store[:last]
    return ret, true

Here, you can't just foo.pop() and pretend that it returns only one value, you have to use the val, _ := foo.pop() semantics. It's the exact same use-case - I should be able to ignore the second return value here if I want to. So why impose this restriction?

I'd love to know what the rationale behind this was.

Date: 2023-02-13

Author: Brian Kamotho