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.