1

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..

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3
  • 1
    Strongly related: unix.stackexchange.com/questions/65803/… Commented Mar 21, 2018 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} Commented Mar 21, 2018 at 19:23
  • methinks /user/ilkkachu has OCD. :) Commented Mar 23, 2018 at 8:33

1 Answer 1

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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 "${test:0:2}"
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2
  • Why then does this have the same problem? > test="-efoo"; answer=${test:0:2}; echo $answer Commented Mar 21, 2018 at 19:20
  • 1
    You’re still passing -e to echo... Commented Mar 21, 2018 at 19:28

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