Character arrays - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is a character array in C?

A character array in C is a sequence of characters stored in contiguous memory locations. It is often used to store strings, which are sequences of characters ending with a special null character '\0'.

beginner

How do you declare a character array to hold the word "hello"?

You can declare it like this:
char word[6] = {'h', 'e', 'l', 'l', 'o', '\0'};
or simply:
char word[] = "hello";
The extra space is for the null character '\0' that marks the end of the string.

beginner

Why do character arrays used as strings in C end with a '\0' character?

The '\0' character, called the null terminator, tells functions where the string ends. Without it, functions like printf or strlen wouldn't know where the string stops and might read garbage data or cause errors.

beginner

How can you change the first character of a character array named 'name' to 'J'?

You can assign it directly like this:
name[0] = 'J';
This changes the first character stored in the array to 'J'.

intermediate

What happens if you forget to add the '\0' at the end of a character array used as a string?

If the '\0' is missing, functions that expect a string will keep reading memory past the array until they find a '\0'. This can cause unexpected behavior, crashes, or security issues.

Which of the following correctly declares a character array to store the string "cat"?

A char animal[3] = {'c', 'a', 't', '\0'};

B char animal[3] = "cat";

C char animal[] = {'c', 'a', 't'};

D char animal[4] = "cat";

Option D is correct because the array size must include space for the null terminator '\0'. So 4 characters are needed for 'c', 'a', 't', and '\0'.

What does the '\0' character represent in a character array?

A The end of the string

B A tab character

C A newline character

D A space character

The '\0' character is the null terminator that marks the end of a string in C.

How do you access the third character in a character array named 'word'?

A word[3]

B word[2]

C word(3)

D word{2}

Arrays in C are zero-indexed, so the third character is at index 2.

What will happen if you print a character array without a '\0' terminator using printf with %s?

A It prints the array correctly

B It prints nothing

C It may print extra garbage characters or crash

D It prints only the first character

Without the '\0', printf doesn't know where the string ends and may print extra memory or cause a crash.

Which of these is a valid way to initialize a character array with the string "dog"?

A char pet[] = "dog";

B char pet[3] = "dog";

C char pet[2] = {'d', 'o', 'g', '\0'};

D char pet = "dog";

Option A is valid because it lets the compiler size the array including the null terminator automatically.

Explain what a character array is and how it is used to store strings in C.

Think about how words are stored as letters in a row with a special end marker.

Describe why the null terminator '\0' is important in character arrays used as strings.

Consider what happens if you don't tell where the string ends.

Practice

1. What is a character array in C used for?

easy

A. To store a sequence of characters, like a word or sentence

B. To store only numbers

C. To perform mathematical calculations

D. To create a list of integers

Solution

1. Step 1: Understand character arrays

   A character array in C is a collection of characters stored in contiguous memory locations.

2. Step 2: Identify its use

   It is mainly used to store text like words or sentences, not numbers or calculations.

3. Final Answer:

   To store a sequence of characters, like a word or sentence -> Option A

4. Quick Check:

   Character array = text storage [OK]

Hint: Character arrays hold text, not numbers or math [OK]

Common Mistakes:

• Thinking character arrays store numbers only
• Confusing character arrays with integer arrays
• Assuming character arrays perform calculations

2. Which of the following is the correct way to declare a character array to hold the word "hello" in C?

easy

A. char word[6] = "hello";

B. char word[5] = {'h', 'e', 'l', 'l', 'o'};

C. char word[] = {'h', 'e', 'l', 'l', 'o', '\0'};

D. char word[5] = "hello";

Solution

1. Step 1: Understand string termination in C

   Strings in C end with a special character '\0' to mark the end.

2. Step 2: Check array size and initialization

   char word[6] = "hello"; declares an array of size 6 (5 letters + '\0') and initializes it with "hello" correctly.

3. Final Answer:

   char word[6] = "hello"; -> Option A

4. Quick Check:

   String needs space for '\0' [OK]

Hint: Remember to include space for '\0' in string arrays [OK]

Common Mistakes:

• Not allocating space for the null terminator '\0'
• Using single quotes for strings
• Declaring array size too small

3. What will be the output of the following C code?

char str[] = "cat";
printf("%c", str[3]);

medium

A. t

B. 0 (null character, no visible output)

C. c

D. Compilation error

Solution

1. Step 1: Understand string indexing

   String "cat" has characters at indices 0:'c', 1:'a', 2:'t', and index 3 is the null terminator '\0'.

2. Step 2: Print character at index 3

   Printing str[3] prints the null character which is invisible, so no visible output appears.

3. Final Answer:

   0 (null character, no visible output) -> Option B

4. Quick Check:

   str[3] = '\0' (null) [OK]

Hint: Remember strings end with '\0' at last index [OK]

Common Mistakes:

• Expecting 't' at index 3 instead of '\0'
• Thinking it causes an error
• Confusing index positions

4. Identify the error in this code snippet:

char name[5];
strcpy(name, "hello world");

medium

A. No error, code works fine

B. Incorrect use of strcpy function

C. Missing #include <string.h>

D. Array size is too small for the string

Solution

1. Step 1: Check array size vs string length

   "hello world" has 11 characters plus '\0' = 12 total, but array size is only 5.

2. Step 2: Understand consequences

   Copying this string will overflow the array, causing undefined behavior.

3. Final Answer:

   Array size is too small for the string -> Option D

4. Quick Check:

   Array must fit string + '\0' [OK]

Hint: Array size must be >= string length + 1 [OK]

Common Mistakes:

• Ignoring array size limits
• Assuming strcpy checks size
• Forgetting null terminator space

5. Given the code below, what will be the output?

char arr[10] = "abc";
arr[3] = 'd';
arr[4] = '\0';
printf("%s", arr);

hard

A. abc\0d

B. abc

C. abcd

D. Runtime error

Solution

1. Step 1: Analyze initial array content

   arr is initialized with "abc" which is 'a','b','c','\0' plus extra space.

2. Step 2: Modify characters and null terminator

   arr[3] = 'd' adds 'd' after 'c', then arr[4] = '\0' sets new string end after 'd'.

3. Step 3: Print the string

   printf prints characters until '\0', so output is "abcd".

4. Final Answer:

   abcd -> Option C

5. Quick Check:

   String ends at arr[4] '\0' [OK]

Hint: Null terminator controls string end in arrays [OK]

Common Mistakes:

• Ignoring manual null terminator placement
• Expecting original string without changes
• Confusing character and string printing