String Basics and Memory Representation in DSA C - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of String Pointer Arithmetic

What is the output of the following C code snippet?

DSA C

char *str = "hello";
printf("%c", *(str + 1));

A e

B h

C o

D l

💡 Hint

Remember that pointer arithmetic moves to the next character in the string.

The pointer 'str' points to the first character 'h'. Adding 1 moves to the second character 'e'. Dereferencing gives 'e'.

❓ Predict Output

intermediate

String Length Calculation Using Pointer

What is the value of 'length' after running this code?

DSA C

char *s = "abcde";
int length = 0;
while (*(s + length) != '\0') {
    length++;
}

A 4

B 5

C 6

D 0

💡 Hint

Count characters until the null terminator '\0'.

The loop increments length until it reaches the null character, counting all 5 characters.

🧠 Conceptual

advanced

Understanding String Memory Layout

Which statement correctly describes how the string "data" is stored in memory in C?

A Characters are stored backwards in memory ending with '\0'.

B Characters are stored in random memory locations with a pointer array pointing to each.

C The string is stored as a single integer representing all characters.

D Each character is stored in consecutive memory locations followed by a null terminator '\0'.

💡 Hint

Think about how arrays and strings are stored in C.

In C, strings are arrays of characters stored consecutively in memory, ending with a null terminator.

🔧 Debug

advanced

Identify the Error in String Modification

What error occurs when running this code snippet?

DSA C

char *str = "hello";
str[0] = 'H';

A Segmentation fault (attempt to modify read-only memory)

B No error, output is 'Hello'

C Compilation error: cannot assign to str[0]

D Runtime error: null pointer dereference

💡 Hint

String literals are stored in read-only memory.

String literals are stored in read-only memory; modifying them causes a segmentation fault.

🚀 Application

expert

Result of Pointer Increment on String Array

Given the code below, what is printed?

DSA C

char arr[] = "abc";
char *p = arr;
printf("%c", *++p);

A \0

B a

C b

D c

💡 Hint

The operator ++p increments the pointer before dereferencing.

The pointer p is incremented to point to 'b', then dereferenced to print 'b'.

Practice

1. Which of the following correctly describes how strings are stored in C?

easy

A. As an array of characters ending with a null character '\0'.

B. As a linked list of characters.

C. As a single integer representing the string length.

D. As a floating-point number encoding characters.

Solution

1. Step 1: Understand C string storage

   In C, strings are stored as arrays of characters with a special null character '\0' marking the end.

2. Step 2: Compare options with this knowledge

   Only As an array of characters ending with a null character '\0'. correctly states this; others describe different data structures or types.

3. Final Answer:

   As an array of characters ending with a null character '\0'. -> Option A

4. Quick Check:

   C strings end with '\0' = A [OK]

Hint: Remember C strings always end with '\0' to mark their end [OK]

Common Mistakes:

• Thinking strings are linked lists in C
• Assuming strings store length explicitly
• Confusing string storage with other data types

2. Which of the following is the correct way to declare and initialize a string literal in C?

easy

A. char str = "hello";

B. char str[5] = 'hello';

C. char str[] = "hello";

D. char str[] = {'h','e','l','l','o'};

Solution

1. Step 1: Check string declaration syntax

   In C, to store a string literal, use an array of chars with double quotes and square brackets.

2. Step 2: Evaluate each option

   A uses char (single char) with double quotes (wrong). B uses single quotes for multiple chars (wrong). D misses null terminator '\0'. C correctly declares and initializes with null terminator.

3. Final Answer:

   char str[] = "hello"; -> Option C

4. Quick Check:

   Use char array with double quotes = C [OK]

Hint: Use char array with double quotes for string literals [OK]

Common Mistakes:

• Using single quotes for strings
• Not including null terminator
• Declaring char variable instead of array

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

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

medium

A. c

B. \0 (null character, prints nothing visible)

C. a

D. Runtime error

Solution

1. Step 1: Understand string indexing and null terminator

   The string "abc" is stored as {'a','b','c','\0'}. Index 3 is the null character '\0'.

2. Step 2: Determine printf output for str[3]

   Printing '%c' with '\0' prints nothing visible (null character), no error occurs.

3. Final Answer:

   \0 (null character, prints nothing visible) -> Option B

4. Quick Check:

   Index 3 is '\0' = A [OK]

Hint: Remember string ends with '\0' at last index + 1 [OK]

Common Mistakes:

• Assuming str[3] is 'a' or 'c'
• Expecting runtime error for accessing null char
• Confusing null char with printable character

4. Identify the error in the following C code snippet:

char str[5] = "hello";
str[5] = '!';

medium

A. Array index out of bounds when assigning str[5].

B. Incorrect use of single quotes for string.

C. Missing null terminator in string initialization.

D. No error, code runs fine.

Solution

1. Step 1: Analyze array size and indexing

   Array str has size 5, valid indices are 0 to 4. Accessing str[5] is out of bounds.

2. Step 2: Check string initialization and assignment

   "hello" fits exactly 5 chars including '\0'? No, 'hello' needs 6 chars including '\0'. But main error is str[5] assignment which is invalid index.

3. Final Answer:

   Array index out of bounds when assigning str[5]. -> Option A

4. Quick Check:

   Index 5 invalid for size 5 array = D [OK]

Hint: Array indices go from 0 to size-1 only [OK]

Common Mistakes:

• Ignoring out of bounds access
• Confusing array size with string length
• Thinking str[5] is valid index

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

char str1[] = "abc";
char *str2 = str1;
str2[1] = 'z';
printf("%s", str1);

hard

A. abc

B. Runtime error

C. a\0c

D. azc

Solution

1. Step 1: Understand pointer and array relation

   str2 points to str1 array. Changing str2[1] changes str1[1].

2. Step 2: Apply the character change and print

   str2[1] = 'z' changes 'b' to 'z'. Printing str1 shows "azc".

3. Final Answer:

   azc -> Option D

4. Quick Check:

   Pointer modifies original array = B [OK]

Hint: Pointer to array modifies original string content [OK]

Common Mistakes:

• Thinking str1 remains unchanged
• Expecting null character insertion
• Assuming runtime error due to pointer