Bash/SH, Same command different output?

Question

$ cat a.sh
#!/bin/bash

echo -n "apple" | shasum -a 256

$ sh -x a.sh
+ echo -n apple
+ shasum -a 256
d9d20ed0e313ce50526de6185500439af174bf56be623f1c5fe74fbb73b60972  -
$ bash -x a.sh
+ echo -n apple
+ shasum -a 256
3a7bd3e2360a3d29eea436fcfb7e44c735d117c42d1c1835420b6b9942dd4f1b  -

And the last one is correct. Why is that? and how to solve it?

For reference, I get the last one if I try csh or zsh also. But I get the same result for all 4 if I omit -n from echo. — SethMMorton
– SethMMorton, Commented Dec 16, 2016 at 3:06
Moral of the story: if the output of the command differs, rethink your assumptions about its input. — chepner
– chepner, Commented Dec 17, 2016 at 15:37

mklement0 · Accepted Answer · 2017-01-17 02:50:43Z

13

Per POSIX, echo supports no options.

Therefore, when echo -n is run with sh, it outputs literal -n instead of interpreting -n as the no-trailing-newline option:

$ sh -c 'echo -n "apple"'
-n apple                  # !! Note the -n at the beginning.

^{Note: Not all sh implementations behave this way; some, such as on Ubuntu (where dash acts as sh), do support the -n option, but the point is that you cannot rely on that, if your code must run on multiple platforms.}

The portable POSIX-compliant way to print to stdout is to use the printf utility:

printf %s "apple" | shasum -a 256

edited Jan 17, 2017 at 2:50

answered Dec 16, 2016 at 3:09

mklement0

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