Why does substring expansion [ test="-efoo"; echo ${test:0:2} ] fail?

Question

On Linux Mint, using bash..

test="-ffoo"; echo ${test:0:2}

works outputting the first two characters

but

test="-efoo"; echo ${test:0:2}

fails, with apparently null output.

I'm thinking the form of this is

${parameter:offset:length}

I know enough that parameter characters cannot be *@#?-$!0_

but $test is the parameter - surely its contents can be anything? I guess -e is triggering something shell-like but why..

Strongly related: https://unix.stackexchange.com/questions/65803/why-is-printf-better-than-echo — Jeff Schaller, Mar 21 '18 at 18:42
This is more about getting the correct answer than the diff between echo and printf. [test="-efoo"; answer=${test:0:2}; echo $answer ] seems to have the same problem with answer not seeing the correct result from ${test:0:2} — David Brown, Mar 21 '18 at 19:23

score 3 · Accepted Answer · edited Mar 22 '18 at 13:58

3

When you run

test="-efoo"; echo ${test:0:2}

echo is run with the argument -e, which in some echo implementations including the echo builtin command of most bash deployments, is a valid option and is thus “swallowed”.

Use printf instead:

test="-efoo"; printf %s\\n "${test:0:2}"

edited Mar 22 '18 at 13:58

Stéphane Chazelas

544,893

answered Mar 21 '18 at 18:43

Stephen Kitt

434,908

Why then does this have the same problem? > test="-efoo"; answer=${test:0:2}; echo $answer – David Brown Mar 21 '18 at 19:20
1

You’re still passing -e to echo... – Jeff Schaller Mar 21 '18 at 19:28

Why does substring expansion [ test="-efoo"; echo ${test:0:2} ] fail?

1 Answers1