bash redirect to the same file in a bad manner

Question

In bash, I know that foo 1>a.txt 2>&1 outputs both stdout and stderr to the same file. But, if I replace this by foo 1>a.txt 2>a.txt , it behaves strangely.

GNU bash, version 5.1.16(1)-release (x86_64-pc-linux-gnu)

A sample code is here:

#!/usr/bin/bash

function foo {
  echo stdout1
  echo stdout2
  echo stdout3
  echo stderr1 >&2
  echo stderr2 >&2
  echo stdout4
}

foo 1>a.txt 2>a.txt

Then, the content of a.txt is

stderr1
stderr2
stdout3
stdout4

A strange point is I would expect that stdout does not appear at all or, behaves like 1> a.txt 2>&1, but it is not so. A problem is that a.txt is open in write mode twice. I want to know what happens in this redirections.

Answer 1

This is an expected behavior but looks strange just because the number of characters of the English words stderr and stdout are same.

The root cause is a.txt is open twice, giving bash two file handles which both point to a.txt. They are completely independent, meaning the seek positions are not synced.

Let's see it step by step. When

echo stdout1
echo stdout2
echo stdout3

is executed, the content of the file becomes

stdout1
stdout2
stdout3

Then

echo stderr1 >&2
echo stderr2 >&2

is executed but the seek position of stderr is still 0 so the first and second line is overwritten, giving

stderr1
stderr2
stdout3

Finally,

echo stdout4

gives

stderr1
stderr2
stdout3
stdout4

because the seek position of stdout is at the start of the 4th line due to the previous execution of echo stdout1; echo stdout2; echo stdout3.