Checking Equality of Three Values Elegantly

While the traditional approach with if a == b == c results in a syntax error, there are alternative methods to determine whether three values are equal.

Using a Clear and Concise Approach

The simplest solution remains:

if a == b && a == c {
    fmt.Println("All 3 are equal")
}

This solution is straightforward and efficient, making comparisons on a per-pair basis.

Exploring Creative Solutions

Using a Map as a Set:

The len() function returns the number of unique keys in a map. By using a map with interface{} keys, we can check if all values are equal by comparing the map length to 1:

if len(map[interface{}]int{a: 0, b: 0, c: 0}) == 1 {
    fmt.Println("All 3 are equal")
}

With Arrays:

Arrays are comparable, allowing us to compare multiple elements at once:

if [2]interface{}{a, b} == [2]interface{}{b, c} {
    fmt.Println("All 3 are equal")
}

Using a Tricky Map:

We can index a map with a key that results in the comparison value:

if map[interface{}]bool{a: b == c}[b] {
    fmt.Println("All 3 are equal")
}

With Anonymous Structs:

Structs are also comparable, so we can create an anonymous struct with the values and compare them:

if struct{ a, b interface{} }{a, b} == struct{ a, b interface{} }{b, c} {
    fmt.Println("All 3 are equal")
}

With Slices:

To compare slices, we utilize the reflect.DeepEqual() function:

if reflect.DeepEqual([]interface{}{a, b}, []interface{}{b, c}) {
    fmt.Println("All 3 are equal")
}

Using a Helper Function:

We can define a helper function to handle any number of values:

func AllEquals(v ...interface{}) bool {
    if len(v) > 1 {
        a := v[0]
        for _, s := range v {
            if a != s {
                return false
            }
        }
    }
    return true
}