How to make echo interpret backslash escapes and not print a trailing newline?

Question

I would like to use `echo` in bash to print out a string of characters followed by only a carriage return. I've looked through the man page and have found that `echo -e` will make `echo` interpret backslash escape characters. Using that I can say `echo -e 'hello\r'` and it will print like this ``` $>echo -e 'hello\r' hello $> ``` So it looks like it handled the carriage return properly. I also found `echo -n` in the man page will stop `echo` from inserting a newline character and it looks like it works when I do this ``` $>echo -n 'hello\r' hello\r$> ``` The problem I'm having is in combining both `-e` and `-n`. I've tried each of `echo -e -n 'hello\r'`, `echo -n -e 'hello\r'`, `echo -en 'hello\r'`, and `echo -ne 'hello\r'` and nothing gets printed like so: ``` $>echo -ne 'hello\r' $> ``` Is there something I'm missing here or can the `-e` and `-n` options not be used together?

Accepted Answer

I think it's working, you're just not seeing it. This is your command:

```
$> echo -ne 'hello\r' 
```

Because of the carriage return (`\r`), that will leave the cursor at the start of the same line on the terminal where it wrote the `hello` - which means that's where the next thing output to the terminal will be written. So if your actual prompt is longer than the `$>` you show here, it will overwrite the `hello` completely.

This sequence will let you see what's actually happening:

```
echo -ne 'hello\r'; sleep 5; echo 'good-bye'
```

But for better portability to other shells, I would avoid using options on `echo` like that. Those are purely bashisms, not supported by the POSIX standard. The `printf` builtin, however, is specified by POSIX. So if you want to display strings with no newline and parsing of backslash sequences, you can just use it:

```
printf '%s\r' 'hello'
```

How to make echo interpret backslash escapes and not print a trailing newline?

Solution

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