Question

* Show the output of the following BASH code:

i=7
i=$((i-5))
if [ $i -eq 4 ] # -eq is ==
then
        echo "Four"
elif [ $i -eq 2 ] then
        echo "Two"
else
        echo $i
fi

Answer 1

Output of the above BASH code is "Two"

Explanation:

In the program "i" value is 7 then $i = $((i-5)) that is 7-5=2

Then the value of $i=2

In the code the if condition is that "if $i = 4 , Then display "Four" "

"if $i=2, Then display "Two" otherwise display value of "$i"

Since the $i value is 2 , the program displays "Two" as output

This is the reason , the program displays "Two" as output.

Error in the code:

In the above code , at elif statement "then" must be in next line

otherwise program will display an error.

Correct code:

i=7
i=$((i-5))
if [ $i -eq 4 ] # -eq is ==
then
        echo "Four"
elif [ $i -eq 2 ]
then
        echo "Two"
else
        echo $i
fi

Screenshot of the running code and Output: