You want to write a script that prints "Even" if a number stored in n is even, and "Odd" otherwise. Which of these is the correct if-then-fi structure to do this?
hard🚀 Application Q15 of 15
Bash Scripting - Conditionals
You want to write a script that prints "Even" if a number stored in n is even, and "Odd" otherwise. Which of these is the correct if-then-fi structure to do this?
Aif [ $((n % 2)) -eq 0 ]; then echo "Even"; else echo "Odd"; fi
Bif [ $n % 2 == 0 ]; then echo "Even"; else echo "Odd"; fi
Cif (( n % 2 == 0 )) then echo "Even"; else echo "Odd"; fi
Dif [ $n / 2 -eq 0 ]; then echo "Even"; else echo "Odd"; fi
Step-by-Step Solution
Solution:
Step 1: Understand how to check evenness in bash
We use arithmetic expansion $(( )) to calculate remainder of n divided by 2. If remainder is 0, number is even.
Step 2: Analyze each option's correctness
if [ $((n % 2)) -eq 0 ]; then echo "Even"; else echo "Odd"; fi correctly uses [ $((n % 2)) -eq 0 ] with then, else, fi. if [ $n % 2 == 0 ]; then echo "Even"; else echo "Odd"; fi uses invalid syntax inside [ ]. if (( n % 2 == 0 )) then echo "Even"; else echo "Odd"; fi misses semicolon before 'then' causing syntax error. if [ $n / 2 -eq 0 ]; then echo "Even"; else echo "Odd"; fi incorrectly uses division instead of modulo.
Final Answer:
if [ $((n % 2)) -eq 0 ]; then echo "Even"; else echo "Odd"; fi -> Option A
Quick Check:
Modulo with $(( )) inside [ ] and proper if-then-fi = A [OK]
Quick Trick:Use $(( )) for math inside [ ] in if-then-fi [OK]
Common Mistakes:
MISTAKES
Using % inside [ ] without $(( ))
Using == instead of -eq for numbers
Confusing division with modulo
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