Hi,
your for loop condition skips odd numbered argument counts. Consider:
- args = ["--bad"]
- len(args) = 1
- for i+1 < len(args) --> condition is false (1 not less than 1)
Thus the only way to trigger the os.Exit(1) is to have at least 2
arguments, of which the first is not "--name" or "--age". For example:
args= ["--bad", "anything"]
However if the bad argument is the last, then the for loop exits early.
-Diego
On Mon, 23 Sept 2024 at 15:42, Maisa Unbelievable <
emilygraceseville7...@gmail.com> wrote:
> Hello there! I don't quiet understand why *os.Exit* doesn't terminate my
> program according to its documentation:
>
> Exit causes the current program to *exit with the given status code*.
>
> My program implements a rudimentary CLI parsing:
>
> package main
>
> import (
> "fmt"
> "os"
> )
>
> func main() {
> var name string
> var age string
>
> args := os.Args[1:]
> i := 0
>
> for i+1 < len(args) {
> option := args[i]
> value := args[i+1]
>
> switch option {
> case "--name":
> name = value
> case "--age":
> age = value
> default:
> fmt.Printf("Unknown option %s", option)
> os.Exit(1) // <-- I expected this function to terminate the program
> }
>
> i += 2
> }
>
> fmt.Printf("name == %s, age == %s", name, age)
> }
>
> Basically, I wanna rewrite this Fish program to Go:
>
> #!/usr/bin/env fish
>
> set --local name
> set --local age
>
> set --local i 1
>
> while set --query argv[$i]
> set --local option "$argv[$i]"
> set --local value "$argv[$(math $i + 1)]"
>
> switch "$option"
> case --name
> set name $value
> case --age
> set age $value
> case '*'
> printf "Unknown option %s" $option
> return 1 # I expected os.Exit to be a "return {{code}}" shell
> equivalent
> end
>
> set i (math $i + 2)
> end
>
> printf "name == %s, age == %s" $name $age
>
> For the context: I'm new to go but have shell scripting experience.
>
> --
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> .
>
--
Diego Joss
Electrical Engineer
SamanTree Medical SA
Avenue de Provence 12, 1007 Lausanne, Switzerland
Tel: +41 21 625 08 97
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