Python zero as file parameter in open function

Question

I'm currently puzzling around this piece of code.

What exactly does

open(0)

do?

I already looked up the docs or tried to find something in the internet but no clues.

The codesnippet where this code is used:

map(abs,map(int,open(0).read().split()))

Thanks ^

Does this answer your question? [Integer File Descriptor "0" in open()](https://stackoverflow.com/questions/53898231/integer-file-descriptor-0-in-open) — Sandeep Gusain, Mar 19 '21 at 20:14

score 6 · Accepted Answer · answered Mar 19 '21 at 20:11

0 is the file descriptor associated with stdin (1 corresponds to stdout, 2 to stderr). open takes int file descriptors as an argument, not just paths, so passing 0 is legal. This is just a somewhat obscure way to create a file object bound to stdin without needing to import sys.

The flaw with it is that, when it is closed, it will close the file descriptor 0 (since closefd=False was not passed), so sys.stdin will be closed without realizing it (though it'll probably figure it out when if someone tries to use it).

So it will just read input, right? Nevermind, got it now – Ari24 Mar 19 '21 at 20:15 — Ari24, Mar 19 '21 at 20:15

Python zero as file parameter in open function

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