[Go]Comma OK Idiom

Have you ever seen Go code like this?

if seconds, ok := timeZone[tz]; ok {
    return seconds
}

This is called “comma ok” idiom.

For obvious reasons this is called the “comma ok” idiom.

via. Effective Go - The Go Programming Language

Simple Map

Let’s take a deeper look.

var timeZone = map[string]int{
        "UTC": 0 * 60 * 60,
        "EST": -5 * 60 * 60,
        "CST": -6 * 60 * 60,
        "MST": -7 * 60 * 60,
        "PST": -8 * 60 * 60,
}
var val1 = timeZone["PST"]
var val2 = timeZone["invalid"]
fmt.Printf("val1: %d val2: %d\n", val1, val2)

The output is:

val1: -28800 val2: 0

If the key is found in map, it returns the value, otherwise, it returns zero value - in this examaple it returns 0.

Multiple Assignment

How about multiple assignment? let’s see an examaple.

var val1, ok1 = timeZone["PST"]
var val2, ok2 = timeZone["invalid"]
fmt.Printf("val1: %d ok1: %t val2: %d ok2: %t\n", val1, ok1, val2, ok2)

The output is:

val1: -28800 ok1: true val2: 0 ok2: false

If the key is found in map, it returns the value with true, otherwise, it returns 0 with false.

comma ok idiom

Let’s back to the first exmaple:

if seconds, ok := timeZone[tz]; ok {
    return seconds
}

This means that if the key(tz) is found in timeZone map, it returns seconds with true(ok).

This idiom is handy when you’d like to configure fallback value:

if seconds, ok := timeZone[tz]; ok {
    return seconds
}

log.Println("unknown time zone:", tz)
return fallbackValue

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