Character arrays - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of character array initialization

What is the output of this C program?

C

#include <stdio.h>

int main() {
    char arr[] = {'H', 'e', 'l', 'l', 'o'};
    printf("%s", arr);
    return 0;
}

A Random characters or garbage output

B Hello\0

C Compilation error

D Hello

💡 Hint

Remember how strings are represented in C and what %s expects.

The array arr is not null-terminated, so printf with %s reads beyond the array until it finds a null character, causing undefined or garbage output.

❓ Predict Output

intermediate

Length of character array with null terminator

What will be the output of this program?

C

#include <stdio.h>
#include <string.h>

int main() {
    char arr[] = "Hello";
    printf("%zu", strlen(arr));
    return 0;
}

A 6

B Undefined behavior

C Compilation error

D 5

💡 Hint

strlen counts characters until the null terminator, but does not include it.

The string "Hello" is stored with a null terminator, so strlen returns 5, the number of visible characters.

🔧 Debug

advanced

Identify the error in character array assignment

What error will this code produce?

C

#include <stdio.h>

int main() {
    char arr[5];
    arr = "Hello";
    printf("%s", arr);
    return 0;
}

A Output: Hello

B Compilation error: array type is not assignable

C Runtime error: segmentation fault

D Compilation error: missing semicolon

💡 Hint

Think about how arrays and pointers work in C.

Arrays cannot be assigned to after declaration. The line arr = "Hello" is invalid and causes a compilation error.

📝 Syntax

advanced

Correct way to declare and initialize a character array

Which option correctly declares and initializes a character array to hold the string "Code" by explicitly including the null terminator?

A char arr[5] = {'C', 'o', 'd', 'e'};

B char arr[] = "Code";

C char arr[5] = {'C', 'o', 'd', 'e', '\0'};

D char arr[4] = {'C', 'o', 'd', 'e'};

💡 Hint

Remember the null terminator is needed for strings in C.

Option C correctly allocates space for 5 characters including the null terminator and initializes the array properly.

🚀 Application

expert

Number of elements in a character array

Given the code below, what is the value of variable count after execution?

C

#include <stdio.h>

int main() {
    char arr[] = "Programming";
    int count = sizeof(arr) / sizeof(arr[0]);
    printf("%d", count);
    return 0;
}

A 12

B 11

C 10

D Undefined behavior

💡 Hint

Remember that string literals include the null terminator in arrays.

The string "Programming" has 11 characters plus 1 null terminator, so the array size is 12 elements.

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