import "a_module_path"
optionally rename it to make it clearer
import util "a_module_path"
- sean
On Sat, Apr 15, 2023, 20:31 Victor Giordano <vitucho3...@gmail.com> wrote:
> Hi there!
>
> I was playing a little bit with modules and packages, regarding making
> projects.
> And I'm struggling to use a package (non-main) declared in the root
> directory.. allow me to show an example:
>
> This scenario, I have project somewhere on my file system
> .
> ├── cmd
> │ └── main.go
> ├── go.mod
> └── util.go
>
> *go.mod*
> module a_module_path
>
> go 1.19
>
>
> *util.go*
> package util
>
> func Abs(x int) int {
> if x < 0 {
> return -x
> } else {
> return x
> }
> }
>
> And the problem appears when I try to use the package util (in the root
> directory of the module) within another package of the module...
>
> *cmd/main.go*
> package main
>
> import (
> "fmt"
> "a_module_path/util" // this doesn't works
> "a_module_path/../util" // this attemp neither (and as relative import
> paths are not supported in module mode i guess is a no-go)
> )
>
> func main() {
> fmt.Println(util.Abs(-2))
> }
>
> The question, as you may predict, is *¿If there any way to make this
> work?*
> *I do not want to create a folder named util (or whatever) and place there
> the util package.*
>
> Thanks for the reading and the patience!
> Keep rocking code!
>
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