Logical operators (-a, -o, !) in Bash Scripting - Practice Problems & Coding Challenges

Challenge - 5 Problems

💻 Command Output

intermediate

What is the output of this Bash test command?

Consider the following Bash command:

if [ 5 -gt 3 -a 2 -lt 4 ]; then echo "True"; else echo "False"; fi

What will be printed when this command runs?

Bash Scripting

if [ 5 -gt 3 -a 2 -lt 4 ]; then echo "True"; else echo "False"; fi

A Syntax error

B True

C False

D No output

💡 Hint

Remember that '-a' means AND in Bash test expressions.

The condition checks if 5 is greater than 3 AND 2 is less than 4. Both are true, so the whole condition is true, printing 'True'.

💻 Command Output

intermediate

What does this Bash condition print?

Look at this Bash snippet:

if [ ! 0 -eq 1 ]; then echo "Yes"; else echo "No"; fi

What will be the output?

Bash Scripting

if [ ! 0 -eq 1 ]; then echo "Yes"; else echo "No"; fi

A Yes

B No

C Syntax error

D No output

💡 Hint

The '!' operator negates the condition that follows it.

0 equals 1 is false, negating it with '!' makes the condition true, so it prints 'Yes'.

💻 Command Output

advanced

What is the output of this complex Bash test?

Analyze this Bash command:

if [ 3 -lt 2 -o \( 4 -eq 4 -a ! 5 -gt 10 \) ]; then echo "Pass"; else echo "Fail"; fi

What will it print?

Bash Scripting

if [ 3 -lt 2 -o \( 4 -eq 4 -a ! 5 -gt 10 \) ]; then echo "Pass"; else echo "Fail"; fi

A No output

B Fail

C Syntax error

D Pass

💡 Hint

Remember '-o' means OR, '-a' means AND, and '!' negates the condition. Parentheses group conditions.

3 < 2 is false, but the grouped condition (4 == 4 AND NOT 5 > 10) is true, so overall condition is true, printing 'Pass'.

💻 Command Output

advanced

What error does this Bash test command raise?

What happens when you run this command?

if [ 5 -gt 3 -a ! ]; then echo "OK"; else echo "Not OK"; fi

Bash Scripting

if [ 5 -gt 3 -a ! ]; then echo "OK"; else echo "Not OK"; fi

A Prints Not OK

B Prints OK

C Syntax error

D No output

💡 Hint

The '!' operator must be followed by a condition.

The '!' operator is used without a condition, causing a syntax error in the test expression.

💻 Command Output

expert

What is the value of variable RESULT after running this script?

Given this Bash script:

RESULT=0
if [ ! \( 2 -eq 2 -o 3 -lt 1 \) ]; then RESULT=1; else RESULT=2; fi
echo $RESULT

What will be printed?

Bash Scripting

RESULT=0
if [ ! \( 2 -eq 2 -o 3 -lt 1 \) ]; then RESULT=1; else RESULT=2; fi
echo $RESULT

A Syntax error

B 1

C 2

D 0

💡 Hint

Evaluate inside the parentheses first, then apply the negation.

Inside parentheses: 2 == 2 is true, 3 < 1 is false, so OR is true. Negating true gives false, so else branch runs, RESULT=2.

Practice

1. Which logical operator in bash scripting means AND when used inside [ ]?

easy

A. &&

B. -o

C. !

D. -a

Solution

1. Step 1: Understand bash logical operators inside [ ]

   Inside single brackets, -a means AND, -o means OR, and ! means NOT.

2. Step 2: Identify the AND operator

   The operator -a is used to combine two conditions with AND logic.

3. Final Answer:

   -a -> Option D

4. Quick Check:

   AND operator = -a [OK]

Hint: Remember: -a is AND, -o is OR, ! is NOT inside [ ] [OK]

Common Mistakes:

• Confusing -a with -o
• Using && inside [ ] which is invalid
• Thinking ! means AND

2. Which of the following is the correct syntax to check if file file.txt exists AND is readable in bash?

easy

A. [ -e file.txt -o -r file.txt ]

B. [ -e file.txt && -r file.txt ]

C. [ -e file.txt -a -r file.txt ]

D. [ ! -e file.txt -a -r file.txt ]

Solution

1. Step 1: Understand the condition to check

   We want to check if file.txt exists AND is readable, so both conditions must be true.

2. Step 2: Choose correct logical operator and syntax

   Inside [ ], -a means AND, so [ -e file.txt -a -r file.txt ] is correct. Using -o means OR, which is wrong here. Using && inside [ ] is invalid syntax.

3. Final Answer:

   [ -e file.txt -a -r file.txt ] -> Option C

4. Quick Check:

   AND check uses -a inside [ ] [OK]

Hint: Use -a for AND inside [ ], not && or -o for AND [OK]

Common Mistakes:

• Using -o instead of -a for AND
• Using && inside [ ] which causes syntax error
• Negating conditions incorrectly with !

3. What will be the output of this script snippet?

if [ -f myfile.txt -o ! -r myfile.txt ]; then echo "Check passed"; else echo "Check failed"; fi

Assuming myfile.txt exists but is NOT readable.

medium

A. Check passed

B. Check failed

C. Syntax error

D. No output

Solution

1. Step 1: Analyze conditions with given file state

   -f myfile.txt is true because file exists. ! -r myfile.txt is true because file is not readable.

2. Step 2: Evaluate OR (-o) logic

   Since OR means either condition true passes, and both are true here, the whole condition is true.

3. Final Answer:

   Check passed -> Option A

4. Quick Check:

   True OR True = True [OK]

Hint: OR (-o) passes if any condition is true [OK]

Common Mistakes:

• Thinking ! -r means readable
• Confusing -o with -a
• Expecting syntax error due to !

4. Identify the error in this bash condition:

if [ !-f myfile.txt -a -r myfile.txt ]; then echo "Valid"; fi

medium

A. Missing spaces around ! operator

B. Incorrect use of -a inside [ ]

C. No error, condition is correct

D. Missing closing bracket ]

Solution

1. Step 1: Check spacing around ! operator

   In bash, ! must have a space after it inside [ ]. The code [ ! -f myfile.txt -a -r myfile.txt ] is correct only if spaces are present. If missing, it causes error.

2. Step 2: Confirm other syntax parts

   The use of -a inside [ ] is valid. The closing bracket is present. So no error there.

3. Final Answer:

   Missing spaces around ! operator -> Option A

4. Quick Check:

   ! needs spaces inside [ ] [OK]

Hint: Always put spaces around ! inside [ ] [OK]

Common Mistakes:

• Using && or || inside [ ]
• Forgetting spaces around operators
• Assuming -a is invalid inside [ ]

5. You want to write a bash condition that checks if a variable var is NOT empty AND file1.txt exists OR file2.txt is readable. Which condition is correct?

hard

A. [ -n "$var" -a ( -e file1.txt -o -r file2.txt ) ]

B. [ -n "$var" -a -e file1.txt -o -r file2.txt ]

C. [ -z "$var" -a -e file1.txt -o -r file2.txt ]

D. [ -n "$var" -o -e file1.txt -a -r file2.txt ]

Solution

1. Step 1: Understand the logic to implement

   We want: var NOT empty AND file1 exists OR file2 readable, i.e., (NOT empty AND file1 exists) OR file2 readable.

2. Step 2: Check each option's logic and syntax

   A: Parentheses for grouping invalid inside [ ], syntax error.
   B: [ -n "$var" -o -e file1.txt -a -r file2.txt ] wrong logic: mixes OR then AND.
   C: [ -z "$var" -a -e file1.txt -o -r file2.txt ] uses -z (empty), wrong for NOT empty.
   D: [ -n "$var" -a -e file1.txt -o -r file2.txt ] correct: left-to-right ((non-empty -a exists) -o readable).

3. Final Answer:

   [ -n "$var" -a -e file1.txt -o -r file2.txt ] -> Option B

4. Quick Check:

   -a/-o left-to-right no grouping needed [OK]

Hint: -a/-o associate left-to-right inside [ ] [OK]

Common Mistakes:

• Trying to group with ( ) invalid in [ ]
• Mixing -o and -a without understanding associativity
• Using -z for non-empty check