grep behaviour is different when run using bash -c '...'

Question

I met an interesting issue while working with this code from Stack Overflow: tripleee's answer on "How to check if a file contains only zeros in a Linux shell?"

Why does the same bash code produce different result depending on interactive shell or subshell?

Make all-zeros file with name your_file.

$ truncate -s 1K your_file

Interactive shell example

$ tr -d '\0' <your_file | grep -am 1 ^ ; echo $?
1

The same code but using subshell

$ bash -c 'tr -d '\0' <your_file | grep -am 1 ^ ; echo $?'

0

And also interesting fact. I changed original example by adding -a option ("equivalent to --binary-files=text") because without this option interactive shell works but subshell:

$ bash -c 'tr -d '\0' <your_file | grep -m 1 ^ ; echo $?'
grep: (standard input): binary file matches
0

P.S. I use bash 5.2.37(1)-release from Ubuntu 25.04

Answer 1

This is just a quoting issue. When you run this:

bash -c 'tr -d '\0' <your_file | grep -am 1 ^ ; echo $?'

It becomes:

1. bash -c 'tr -d '
2. \0 (which is just 0)
3. ' <your_file | grep -am 1 ^ ; echo $?'

You are using single quotes within the single quotes, so your '\0' actually becomes \0 which is just 0. You can see it happen with set -x:

$ bash -c 'tr -d '\0' <your_file | grep -am 1 ^ ; echo $?'
+ bash -c 'tr -d 0 <your_file | grep -am 1 ^ ; echo $?'

0

The presence of NUL (\0) in a file will mark it as binary data. So the fact that grep complained about that is also a hint. As you can see above, you are not actually removing \0, you are removing 0.

Next, the exit status of grep, which is what you have in your $? variable, is 0 if at least one line matches and 1 if no line matches. In your first command, tr -d '\0' <your_file | grep -am 1 ^ ; echo $?, you are deleting all \0 from a file that consists of nothing but \0 so you end up with an empty file. Therefore, nothing is matched and you get an exit status of 1.

With your second command, bash -c 'tr -d '\0' <your_file | grep -am 1 ^ ; echo $?', because of the quoting issue, you are removing 0. The file has no 0, so nothing is removed and there is one "line" to be found and so the output is 0.

The second shell is completely irrelevant, you can get the same behavior without it:

$ tr -d '\0' <your_file | grep -am 1 ^ ; echo $?
1

$ tr -d 0 <your_file | grep -am 1 ^ ; echo $?

0

To get the expected output in the second shell, use:

$ bash -c 'tr -d "\0" <your_file | grep -am 1 ^ ; echo $?'
1