Why Strings Are a Data Structure Not Just Text in DSA C - Test Your Knowledge

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a string variable in C.

DSA C

char name[20] = [1];

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

A'Hello'

B20

CHello

D"Hello"

Attempts:

3 left

💡 Hint

Common Mistakes

Using single quotes for strings

Not using quotes at all

Using a number instead of a string

Strings in C are enclosed in double quotes. Single quotes are for single characters.

2fill in blank

medium

Complete the code to print the string stored in variable 'text'.

DSA C

printf([1], text);

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

A%d

B%c

C%s

D%f

Attempts:

3 left

💡 Hint

Common Mistakes

Using %d or %c instead of %s

Forgetting the format specifier

Use %s to print strings in C with printf.

3fill in blank

hard

Fix the error in the code to copy string 'source' to 'destination'.

DSA C

strcpy(destination, [1]);

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

Adest

Bsource

C&source

D*source

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping source and destination

Using & or * with source incorrectly

strcpy copies from source to destination, so source must be the second argument.

4fill in blank

hard

Fill both blanks to check if the first character of 'word' is 'A'.

DSA C

if (word[1] == [2]) {
    // do something
}

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

A[0]

Bword[0]

C'A'

D"A"

Attempts:

3 left

💡 Hint

Common Mistakes

Using double quotes for characters

Using the whole string instead of first character

Use word[0] to access the first character and compare it with the character 'A' using single quotes.

5fill in blank

hard

Fill all three blanks to create a loop that prints each character of 'str' until the null character.

DSA C

for (int i = 0; [1]; i++) {
    if (str[i] == [2]) break;
    printf("%c", [3]);
}

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

Astr[i] != '\0'

B'\0'

Cstr[i]

Di < strlen(str)

Attempts:

3 left

💡 Hint

Common Mistakes

Using double quotes for '\0'

Using wrong loop condition

Printing the wrong variable

The loop continues while str[i] is not the null character '\0'. We break if we find '\0'. We print str[i] as a character.

Practice

1. Why are strings considered a data structure and not just plain text?

easy

A. Because strings store characters in a sequence that can be accessed and manipulated.

B. Because strings only contain letters and no other data.

C. Because strings are always immutable and cannot be changed.

D. Because strings are only used for displaying text on screen.

Solution

1. Step 1: Understand what a data structure is

   A data structure organizes data so it can be accessed and changed efficiently.

2. Step 2: Analyze string properties

   Strings store characters in order and allow access to each character by position, making them a data structure.

3. Final Answer:

   Because strings store characters in a sequence that can be accessed and manipulated. -> Option A

4. Quick Check:

   Strings = sequence + access [OK]

Hint: Strings hold ordered characters, not just text [OK]

Common Mistakes:

• Thinking strings only hold letters
• Believing strings cannot be changed
• Assuming strings are only for display

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

easy

A. char str = 'hello';

B. char str[] = "hello";

C. string str = "hello";

D. char str = "hello";

Solution

1. Step 1: Recall C string declaration syntax

   In C, strings are arrays of characters ending with a null character '\0'.

2. Step 2: Check each option

   char str[] = "hello"; declares a char array initialized with a string literal correctly.

3. Final Answer:

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

4. Quick Check:

   C strings = char array + null terminator [OK]

Hint: Use char array with double quotes for strings in C [OK]

Common Mistakes:

• Using single quotes for strings
• Using 'string' type which doesn't exist in C
• Declaring char variable for whole string

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

char str[] = "abc";
printf("%c", str[1]);

medium

A. b

B. a

C. c

D. Compilation error

Solution

1. Step 1: Understand string indexing in C

   Strings are arrays; indexing starts at 0, so str[0] = 'a', str[1] = 'b'.

2. Step 2: Identify printed character

   printf prints str[1], which is 'b'.

3. Final Answer:

   b -> Option A

4. Quick Check:

   Index 1 of "abc" = 'b' [OK]

Hint: Remember C arrays start at index 0 [OK]

Common Mistakes:

• Confusing index 1 with index 0
• Expecting whole string output
• Thinking code causes error

4. Find the error in this C code snippet that tries to print the length of a string:

char str[] = "hello";
int len = sizeof(str);
printf("%d", len);

medium

A. printf format specifier %d is wrong for int

B. str is not initialized properly

C. sizeof returns size in bytes including null, not string length

D. No error, code works correctly

Solution

1. Step 1: Understand sizeof on arrays

   sizeof(str) returns total bytes including the null character, not the string length.

2. Step 2: Difference between sizeof and strlen

   strlen(str) returns length excluding null; sizeof includes it, so result differs.

3. Final Answer:

   sizeof returns size in bytes including null, not string length -> Option C

4. Quick Check:

   sizeof != string length [OK]

Hint: Use strlen() for string length, not sizeof() [OK]

Common Mistakes:

• Using sizeof to get string length
• Ignoring null terminator in size
• Assuming code prints length correctly

5. Given the string char str[] = "data";, which code snippet correctly reverses the string in place?

hard

A. str = "atad";

B. for(int i = 0; i < 4; i++) { str[i] = str[3 - i]; }

C. int i = 0; while(str[i] != '\0') { str[i] = str[i+1]; i++; }

D. int i = 0, j = 3; while(i < j) { char temp = str[i]; str[i] = str[j]; str[j] = temp; i++; j--; }

Solution

1. Step 1: Understand in-place reversal logic

   Swapping characters from start and end moving towards center reverses the string.

2. Step 2: Analyze each option

   int i = 0, j = 3; while(i < j) { char temp = str[i]; str[i] = str[j]; str[j] = temp; i++; j--; } swaps characters correctly using two pointers; others overwrite or assign incorrectly.

3. Final Answer:

   int i = 0, j = 3; while(i < j) { char temp = str[i]; str[i] = str[j]; str[j] = temp; i++; j--; } -> Option D

4. Quick Check:

   Swap ends moving inward reverses string [OK]

Hint: Swap characters from ends moving inward [OK]

Common Mistakes:

• Overwriting characters without swapping
• Assigning string literal to array variable
• Ignoring null terminator in loops