compare array in bash

Question

Question: what does this line do? Can someone please explain?

 if [ $((${array[$i]}+1)) -ne ${array[$(($i + 1))]} ] ;then foo; fi

same as if ((array[i] + 1 != array[i+1])); then foo; fi (as long as $i and array[i+1] contain decimal integers). — Stéphane Chazelas
– Stéphane Chazelas, Commented Aug 4, 2016 at 8:49
It produces the value at position [$i] in $array, adds one to it - inside an arithmetic context, then returns (via command substitution) that as the LHS operand to the -ne conditional operator, then evaluates the RHS operand, $(($i + 1)) is variable $i + 1 then the result of this arithmetic operation is used as the key/position in the array ${array[]} - if the LHS and the RHS results are -ne "not equal` the then block runs — the_velour_fog
– the_velour_fog, Commented Aug 4, 2016 at 8:57
And also the same as if [[ array[i]+1 -ne array[i+1] ]] ;then foo; fi as both sides of an integer operator are processed as arithmetic expansions. In math: test for A[i]+1 = A[i+1]. Or, in words: is the next array element the consecutive numeric value of the present one? — user79743
– user79743, Commented Aug 4, 2016 at 11:46

ilkkachu · Accepted Answer · 2016-08-04 08:56:06Z

6

[ ... -ne ... ] - test for inequality
$(( ... + 1)) - add one, arithmetic expansion
${array[$i]} - reference to element $i of an array variable

Or in other words the test is A[i] + 1 != A[i + 1]

answered Aug 4, 2016 at 8:56

ilkkachu

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