If you are fine with cwd, you can use std.fs.cwd().makePath("foobar")
#MakePath
proven needle
you might want makeOpenPath if you want the handle to it immediately as well
rugged bolt
you might want makeOpenPath if you want the handle to it immediately as well
yeah I saw that
rugged bolt
If you are fine with cwd, you can use `std.fs.cwd().makePath("foobar")`
no, not an option. I would much rather supply a path like I have written above
proven needle
std.fs.cwd().makePath() will work with both relative and absolute paths
so even if you give it like C:/foobar it will work
rugged bolt
but cwd() means current working directory?
proven needle
yeah but it still works with absolute paths as well
rugged bolt
yeah but it still works with absolute paths as well
the makePath function for the subpath? thats weird
proven needle
you could also use std.posix.mkdir but both work anyways
the std.fs.openFileAbsolute for example looks like this:
https://github.com/ziglang/zig/blob/master/lib/std/fs.zig#L278-L281
(Simply calls cwd().openFile)
rugged bolt
I didnt find anything in std, is there a function to extract from a path string the filename?
does linux not make this difficult since files can oftentimes have no file extension?
seems like a big os design flaw to me
patent hemlock
*nix extensively uses file signatures -- the first few bytes in the file (e.g. shebang). Mac OS used to rely on the data fork of a file. File extensions are too constrained a namespace for many purposes. All choices have tradeoffs.
rugged bolt
seems like a big tradeoff to me, cant easily judge from a pathname alone if its just a path or path+file
patent hemlock
You still can't even with extensions. On Windows, there's nothing that prevents a folder from having an extension.
