CODEWITHSUNDEEP
go
softshipper
softshipper

Reputation: 34081

It is a type converter to int?

I have a code sample from a book:

The Way To Go: A Thorough Introduction To The Go Programming Language

from which I could not figure out how something works. Look at the code:

package main

import (
    "fmt"
)

type Any interface{}
type EvalFunc func(Any) (Any, Any)

func main() {
    evenFunc := func(state Any) (Any, Any) {
        os := state.(int)
        ns := os + 2
        return os, ns
    }
    even := BuildLazyIntEvaluator(evenFunc, 0)

    for i := 0; i < 10; i++ {
        fmt.Printf("%vth even: %v\n", i, even())
    }
}

func BuildLazyEvaluator(evalFunc EvalFunc, initState Any) func() Any {
    retValChan := make(chan Any)
    loopFunc := func() {
        var actState Any = initState
        var retVal Any
        for {
            retVal, actState = evalFunc(actState)
            retValChan <- retVal
        }
    }
    retFunc := func() Any {
        return <-retValChan
    }
    go loopFunc()
    return retFunc
}

func BuildLazyIntEvaluator(evalFunc EvalFunc, initState Any) func() int {
    ef := BuildLazyEvaluator(evalFunc, initState)
    return func() int {
        return ef().(int)
    }
}

Look at the code line:

return ef().(int)

What's happening here? Does the compiler convert the result into an int type?

Upvotes: 0

Views: 373

Answers (2)

bcmills
bcmills

Reputation: 5187

It's a type assertion - see the Go Spec.

Upvotes: 1

Dave Cheney
Dave Cheney

Reputation: 5755

x := ef()     // x is of type Any, which is actually an interface{}
y := x.(int)  // this is a type assertion, if the contents of x are an interface, y will be assigned x's int value, otherwise the runtime will panic.

Upvotes: 6

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