Terminal - Why the exit command of grep is 0 even if a match is not found?

Question

I have this command: ``` grep -E '^nothing' List.txt | echo $? ``` Here grep doesn't match anything and I simply output its exit code. According to documentation of grep: Normally the exit status is 0 if a line is selected, 1 if no lines were selected, and 2 if an error occurred. However, if the -q or --quiet or --silent option is used and a line is selected, the exit status is 0 even if an error occurred. Other grep implementations may exit with status greater than 2 on error. But: ``` prompt:user$ grep -E '^nothing' List.txt | echo $? 0 prompt:user$ ``` But why do I get 0 as output even if the match doesn't exist, should't I get the expected 1 exit code?

Accepted Answer

This is the problem:

```
grep -E '^nothing' List.txt | echo $?
```

By using single `|` you are sending output of `grep` to `echo` which will always print exit status of previous command and that will always be 0 whether pattern is found or not.

You can use `grep -q`:

```
grep -qE '^nothing' List.txt
```

As per `man grep`:

```
 -q, --quiet, --silent
         Quiet mode: suppress normal output.  grep will only search a file until a match
         has been found, making searches potentially less expensive.
```

Terminal - Why the exit command of grep is 0 even if a match is not found?

Solution

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