19

I have a package named "seeder":

package seeder

import "fmt"

func MyFunc1() {
    fmt.Println("I am Masood")
}

func MyFunc2() {
    fmt.Println("I am a programmer")
}

func MyFunc3() {
    fmt.Println("I want to buy a car")
}

Now I want to call all functions with MyFunc prefix

package main

import "./seeder"

func main() {
    for k := 1; k <= 3; k++ {
        seeder.MyFunc1() // This calls MyFunc1 three times
    }
}

I want something like this:

for k := 1; k <= 3; k++ {
    seeder.MyFunc + k ()
}

and this output:

I am Masood
I am a programmer
I want to buy a car

EDIT1: In this example, parentKey is a string variable which changed in a loop

for parentKey, _ := range uRLSjson{ 
    pppp := seeder + "." + strings.ToUpper(parentKey)
    gorilla.HandleFunc("/", pppp).Name(parentKey)
}

But GC said:

use of package seeder without selector

icza
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Ario
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  • Hi Masood. In your sub functions don't you mean to say the Hello World text matches the function not all MyFunc1?: i.e. func MyFunc2() { fmt.Println("Hello MyFunc2") } – micstr May 23 '16 at 07:04
  • Ok great I see you have fixed this. – micstr May 23 '16 at 07:06
  • Hi no I mean call funcs with special prefix. Lets change fmt output to be more than clear – Ario May 23 '16 at 07:07

2 Answers2

27

You can't get a function by its name, and that is what you're trying to do. The reason is that if the Go tool can detect that a function is not referred to explicitly (and thus unreachable), it may not even get compiled into the executable binary. For details see Splitting client/server code.

With a function registry

One way to do what you want is to build a "function registry" prior to calling them:

registry := map[string]func(){
    "MyFunc1": MyFunc1,
    "MyFunc2": MyFunc2,
    "MyFunc3": MyFunc3,
}
for k := 1; k <= 3; k++ {
    registry[fmt.Sprintf("MyFunc%d", k)]()
}

Output (try it on the Go Playground):

Hello MyFunc1
Hello MyFunc2
Hello MyFunc3

Manual "routing"

Similar to the registry is inspecting the name and manually routing to the function, for example:

func callByName(name string) {
    switch name {
    case "MyFunc1":
        MyFunc1()
    case "MyFunc2":
        MyFunc2()
    case "MyFunc3":
        MyFunc3()
    default:
        panic("Unknown function name")
    }
}

Using it:

for k := 1; k <= 3; k++ {
    callByName(fmt.Sprintf("MyFunc%d", k))
}

Try this on the Go Playground.

Note: It's up to you if you want to call the function identified by its name in the callByName() helper function, or you may choose to return a function value (of type func()) and have it called in the caller's place.

Transforming functions to methods

Also note that if your functions would actually be methods of some type, you could do it without a registry. Using reflection, you can get a method by name: Value.MethodByName(). You can also get / enumerate all methods without knowing their names using Value.NumMethod() and Value.Method() (also see Type.NumMethod() and Type.Method() if you need the name of the method or its parameter types).

This is how it could be done:

type MyType int

func (m MyType) MyFunc1() {
    fmt.Println("Hello MyFunc1")
}

func (m MyType) MyFunc2() {
    fmt.Println("Hello MyFunc2")
}

func (m MyType) MyFunc3() {
    fmt.Println("Hello MyFunc3")
}

func main() {
    v := reflect.ValueOf(MyType(0))
    for k := 1; k <= 3; k++ {
        v.MethodByName(fmt.Sprintf("MyFunc%d", k)).Call(nil)
    }
}

Output is the same. Try it on the Go Playground.

icza
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  • Thank but is there any other way without map? – Ario May 23 '16 at 07:13
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    @MasoodAfrashteh You may do it with a registry, or by transforming functions to methods. Having functions and no registry - you can't. – icza May 23 '16 at 07:16
  • or `reflect`, which helping turn string to function. – Jiang YD May 23 '16 at 07:19
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    @JiangYD You can't get a function by name using `reflect`, only methods. – icza May 23 '16 at 07:22
  • @MasoodAfrashteh Considering your edit to the question, the answers still stand: you can't do it without a registry or using methods instead of functions. – icza May 23 '16 at 07:23
  • Thanks I just change MyType type to string It's working. – Ario May 23 '16 at 07:30
  • yes I missed it. why golang designed not get to `Package` at runtime? – Jiang YD May 23 '16 at 07:32
  • Sorry It's not working because in my example functions write by user and I'm going to connect those functions with special prefix or suffix like "route" to gorilla. – Ario May 23 '16 at 07:43
  • @MasoodAfrashteh As said previously, if they remain functions (and you can't make them methods), you can only do it with a function registry (or some other construct, explicitly listing the names, routing them to functions manually). – icza May 23 '16 at 07:47
1

Another alternative would be to range over an array of your functions

package main

import (
    "fmt"
)

func MyFunc1() {
    fmt.Println("I am Masood")
}

func MyFunc2() {
    fmt.Println("I am a programmer")
}

func MyFunc3() {
    fmt.Println("I want to buy a car")
}


func main() {
   for _, fn := range []func(){MyFunc1, MyFunc2, MyFunc3} {
        fn()
    }
}
Liyan Chang
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