String Basics and Memory Representation in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a string variable in C.

DSA C

char name[] = [1];

Drag options to blanks, or click blank then click option'

A"Hello"

B'Hello'

CHello

D"H"

Attempts:

3 left

💡 Hint

Common Mistakes

Using single quotes for strings

Not using quotes at all

In C, strings are arrays of characters enclosed in double quotes. Single quotes are for single characters.

2fill in blank

medium

Complete the code to print the first character of the string.

DSA C

printf("%c", name[1]);

Drag options to blanks, or click blank then click option'

A(0)

B[1]

C[0]

D.0

Attempts:

3 left

💡 Hint

Common Mistakes

Using parentheses instead of brackets

Starting index at 1

String indexing in C starts at 0, so name[0] gives the first character.

3fill in blank

hard

Fix the error in the code to copy one string to another.

DSA C

strcpy(dest, [1]);

Drag options to blanks, or click blank then click option'

A&source

Bdest

C*source

Dsource

Attempts:

3 left

💡 Hint

Common Mistakes

Passing destination as source

Using & or * operators incorrectly

The source string must be passed as a pointer to strcpy, not the destination or address operators.

4fill in blank

hard

Fill both blanks to declare and initialize a string pointer.

DSA C

char *ptr = [1];

Drag options to blanks, or click blank then click option'

A'World'

B"World"

Cptr

DWorld

Attempts:

3 left

💡 Hint

Common Mistakes

Using single quotes for strings

Not using quotes at all

A string pointer can be initialized to a string literal enclosed in double quotes.

5fill in blank

hard

Fill all three blanks to print the length of a string using strlen.

DSA C

int len = strlen([1]);
printf("Length: %[2]\n", [3]);

Drag options to blanks, or click blank then click option'

Astr

Bd

Clen

Dstrcpy

Attempts:

3 left

💡 Hint

Common Mistakes

Using %c instead of %d for printing length

Passing wrong variable to strlen

strlen takes the string variable, printf uses %c for char, but for int length we use %d. Here %c is incorrect, correct is %d.

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