I want to access a private function of a package called "pastry". But it generates error as: invalid reference to unexported identifier
Specify the way by which private functions of golang are accessible in main.
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6You can't. That's the way the language works.Mr_Pink– Mr_Pink2016-03-15 11:29:41 +00:00Commented Mar 15, 2016 at 11:29
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7The whole idea of "private" is to be non-accessible.Volker– Volker2016-03-15 11:35:41 +00:00Commented Mar 15, 2016 at 11:35
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1Actually it is possible, see my answer.graywolf– graywolf2020-03-11 18:30:11 +00:00Commented Mar 11, 2020 at 18:30
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4 Answers
You can use go:linkname to map functions from same/different package
to some function in yours. For example like:
package main
import (
"fmt"
_ "net"
_ "unsafe"
)
//go:linkname lookupStaticHost net.lookupStaticHost
func lookupStaticHost(host string) []string
func main() {
fmt.Println(lookupStaticHost("localhost"))
}
will produce [127.0.0.1 ::1] when executed on my machine.
Private functions are, by definition, not accessible outside the package in which they are declared.
If you need that function outside that package, you must make it public (change the function name, turning the first letter in upper case).
Eg: if you have func doSomething() rename it to func DoSomething() and use it outside the package with <package name>.DoDomething()
Comments
You can also add public proxy-function.
For example:
You have package-private function
func foo() int {
return 42
}
You can create public function in the same package, which will call the package-private function and return it's result
func Bar() int {
return foo()
}
Comments
in the package (let's say mypackage) where you have pastry function add:
var Pastry = pastry
in main package:
mypackage.Pastry()
1 Comment
typetetris
You are assuming, that he approves altering the source of the package.