4 ms·

I made one of these too! I decided not to use // because I use gofmt auto formatting in my editor and it puts a space between the // and the usr. This one isn't

by fsmv 9mo ago

I made one of these too! I decided not to use // because I use gofmt auto formatting in my editor and it puts a space between the // and the usr. This one isn't changed by gofmt: /*?sr/bin/env go run "$0" "$@"; exit $? #*/

tandr 9mo ago
It works, but the best in me I cannot explain fully first 3 symbols. /*?sr/bin/env finds /usr by expanding *? to a first matching directory. But why not just /*usr/ instead?
- fsmv 9mo ago
  I think I was just trying to minimize accidentally matching the wrong thing. Both do work though and it is kinda nice to be more readable. If I remember right I think ? Is exactly one character only, or maybe it does non greedy .
k776 9mo ago
I wonder if bash has a way to override the default function calling mechanism. Since functions in bashrc would probably take priority, maybe something like this could be possible: (psuedo code) ``` execve(path, args) { if path.endsWith('.go') { go run path args } else { super($path, *args) } } ``` Then when you run `./script.go`, the function gets called and passes it so `go run`, while everything else goes through the existing execve function. That would be interesting, because then you could do this with absolutely anything. `./script.rb` `./script.py` `./script.c` etc
beckit 9mo ago
This is a sweet idea, thank you! My suggestion for minimal/readable: /*usr/bin/env go run "$0" "$@"; exit;*/ - go fmt leaves it alone - it preserves the exit code in case env or go breaks - it "figures out" where go is, no need to hard code a path that breaks on yet another machine (where /usr/bin/env is "standard") ( The first example: - I'm not a big fan of the subtle extra work that the question mark in the original path imposes on the system. env is (almost) always going to be as /usr/bin/env and rarely would something else that matches exist. ls -d /?sr to see what outputs on your system. That being said, the extra work isn't very meaningful in this case... - The trailing $? seems unnecessary as the final exit will convey the return code from env go regardless of if the $? is present or not )

by fsmv 9mo ago

tandr 9mo ago

It works, but the best in me I cannot explain fully first 3 symbols. /*?sr/bin/env finds /usr by expanding *? to a first matching directory. But why not just /*usr/ instead?

fsmv 9mo ago
I think I was just trying to minimize accidentally matching the wrong thing. Both do work though and it is kinda nice to be more readable. If I remember right I think ? Is exactly one character only, or maybe it does non greedy .

k776 9mo ago

I wonder if bash has a way to override the default function calling mechanism. Since functions in bashrc would probably take priority, maybe something like this could be possible: (psuedo code) ``` execve(path, args) { if path.endsWith('.go') { go run path args } else { super($path, *args) } } ``` Then when you run `./script.go`, the function gets called and passes it so `go run`, while everything else goes through the existing execve function. That would be interesting, because then you could do this with absolutely anything. `./script.rb` `./script.py` `./script.c` etc

beckit 9mo ago

This is a sweet idea, thank you! My suggestion for minimal/readable: /*usr/bin/env go run "$0" "$@"; exit;*/ - go fmt leaves it alone - it preserves the exit code in case env or go breaks - it "figures out" where go is, no need to hard code a path that breaks on yet another machine (where /usr/bin/env is "standard") ( The first example: - I'm not a big fan of the subtle extra work that the question mark in the original path imposes on the system. env is (almost) always going to be as /usr/bin/env and rarely would something else that matches exist. ls -d /?sr to see what outputs on your system. That being said, the extra work isn't very meaningful in this case... - The trailing $? seems unnecessary as the final exit will convey the return code from env go regardless of if the $? is present or not )