Can we have function pointers in Google Go?

Question

I was learning about pointers in Google Go. And managed to write something like,

func hello(){

       fmt.Println("Hello World")
}

func main(){

       pfunc := hello     //pfunc is a pointer to the function "hello"
       pfunc()            //calling pfunc prints "Hello World" similar to hello function
}

Is there a way to declare the function pointer without defining it as done above? Can we write something like we do in C?

e.g. void (*pfunc)(void);

Answer 1

It might be possible, although my googling didn't reveal any obvious mechanism for it. It would, however, be a really, really bad idea. You have first-class functions that interact nicely with the type system. They can do everything that function pointers can do except for causing horrendous crashes.

I notice your code comment says that "pfunc is a pointer to the function 'hello'". This is the wrong way to think about first-class functions, even if that might be the implementation mechanism behind the scenes. What that statement really says is "pfunc holds the value that is the function 'hello'". Functions are values just like integers or strings are!

Answer 2

It works if you're using the signature. There's no pointer.

type HelloFunc func(string)

func SayHello(to string) {
    fmt.Printf("Hello, %s!\n", to)
}

func main() {
    var hf HelloFunc

    hf = SayHello

    hf("world")
}

Alternatively you can use the function signature directly, without declaring an own type.

Answer 3

Go doesn't allow function pointers in the same way C and C++ do. There's a pretty good explanation for that on the Go blog.

Here's an example I wrote. Notice how the fp parameter is defined in calculate().

package main

import "fmt"

func main() {
    calculate(Plus)
    calculate(Minus)
    calculate(Multiply)
}

func calculate(fp func(int, int)int) {
    ans := fp(3,2)
    fmt.Printf("\n%v\n", ans) 
}

func Plus(a, b int) int {
    return a + b
}

func Minus(a, b int) int {
    return a - b
}

func Multiply(a,b int) int {
    return a * b
}

The f parameter is defined as as function that takes two ints and returns a single int. This is somewhat the same thing Mue mentioned but shows a different usage example.