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Need help to this bash/shell exercise

JavaScript

The output of this line if expr $x+1 1> /dev/null 2>&1 I really don’t figure out how things go there…

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Answer

Modify your script like this (simple add spaces around the +):

JavaScript

expr tries to add 1 to the value of the variable. If the status is 0 (value of $?) it is a number. Otherwise, there is an error, so it returns 1 (value of $? again).

If you remove the redirections of the output (1>/dev/null 2>&1) you will see that expr e + 1 outputs expr: non-integer argument.

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