How can I "go run" a project with multiple files in the main package?

I have a single file in the main package called main.go. Because the code isn't reusable I want to separate part of the code in a different file but in the same package.

How do I split the contents of main.go into multiple files without creating a separate package?

I want a directory structure like this:

ls foo

# output:
main.go
bar.go

• File: bar.go

package main

import "fmt"

func Bar() {
  fmt.Println("Bar")
}

• File: main.go

package main

func main() {
  Bar()
}

When I run go run main.go, it gives me:

# command-line-arguments
./main.go:4:2: undefined: Bar

Update 26th July 2019 (for go >=1.11)

go run .

Will work on windows as well.

Original answer (for non windows environments)

The code actually works. The problem was that instead of running go run main.go I should run:

go run *.go

Update August 2018, with Go 1.11, a section "Run" states:

The go run command now allows a single import path, a directory name or a pattern matching a single package.
This allows go run pkg or go run dir, most importantly go run .

Original answer Jan. 2015

As mentioned in "How to compile Go program consisting of multiple files?", go run expects a list of files, since it "compiles and runs the main package comprising the named Go source files".
So you certainly can split your main package in several files with go run.

That differs from go build/go install which expect package names (and not go filenames).
A simple go build would produce an executable named after the parent folder.

Note that, as illustrated by this thread, a go run *.go wouldn't work in a Windows CMD session, since the shell doesn't do wildcard expansion.

In my opinion, the best answer to this question is hidden in the comments to the top answer.

Just run this:

go run .

This will run all the files in main package, but will not give an error message like

go run: cannot run *_test.go files (main_test.go)

Kudos to @BarthesSimpson