Variable shadowing in Go

Question

The following code generates the compile error : \"err declared and not used\". If there is a scope/shadowing issue here, it\'s due to a principle I don\'t understand. Can s

Answer 1

This is not because of any shadowing. You have not used err variable that is declared for anything but assigning a value to it .

according to FAQ

The presence of an unused variable may indicate a bug, while unused imports just slow down compilation. Accumulate enough unused imports in your code tree and things can get very slow. For these reasons, Go allows neither

If you declare a variable it has to be used

In the given program err is declared and is being used to assign data to .The value of err is not used at all

You may bipass this kind of error by doing a _ assignment

ie,

  var _ = err

or using err like

  if err != nil {
      fmt.Println(err.Error())
      return
  }

The following code would solve it,but i would suggest use the err for checking error

package main

import (
    "fmt"
)

func main() {
    var (
        err error
        dto = make(map[string]interface{})
    )
    _ = err

    dto[`thing`], err = getThings()
    fmt.Println(dto[`thing`])
}

func getThings() (string, error) {
    return `the thing`, nil
}

PS : You must use variables you declare inside functions, but it's OK if you have unused global variables. It's also OK to have unused function arguments.