expose a function in go package

Question

I would like to expose a function directly from a package.
So I could call my package directly instead of mypackage.Somepublic() method.

package main

import (
    "mypackage"
    "fmt"
)

func main() {
    var result = mypackage() 
    fmt.Println(result)    
}

In node.js, for example, you can expose a anonymous function

module.export = function() {
    console.log('ta da!');
}

or an object

module.export = {  
    doMagic: function() {  
        console.log('ta da!');
    }
};

Answer 1

While there is no direct analogy for your Node.js example, what you can do in Go is something called a "local import." Basically, a local import imports all of the items - functions, types, variables, etc - that a package exports into your local namespace so that they can be accessed as if they had been defined locally. You do this by prefacing the package name with a dot. For example:

import . "fmt"

func main() {
    Println("Hello!") // Same as fmt.Println("Hello!")
}

(See this in action).

This will work for any item that fmt exports. You could do a similar thing with mypackage (this is modified from the code you posted):

package main

import (
    . "mypackage"
    "fmt"
)

func main() {
    var result = Somepublic() // Equivalent to mypackage.Somepublic()
    fmt.Println(result)    
}

Answer 2

Quoting the answer of mko here for better visibility:

Dear Lonely Anonymous adventurer. I guess you got here for the same reason I did. So, the basic answer is. Use capital letter in the name ;) source: tour.golang.org/basics/3 - yes, I was surprised as well. I was expecting some fancy stuff, but that's all it is. Capital letter - export, small letter - no export ;)