How Strings Work Differently Across Languages in DSA C - Code Step by Step

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a string in C.

DSA C

char str[] = [1];

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

A"Hello";

B'Hello'

CHello

D"Hello"

Attempts:

3 left

💡 Hint

Common Mistakes

Using single quotes for strings

Not enclosing the string in quotes

Adding a semicolon inside the quotes

In C, strings are declared as arrays of characters enclosed in double quotes.

2fill in blank

medium

Complete the code to print the string using printf.

DSA C

printf([1], str);

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

A%d

B%s

C%c

D%f

Attempts:

3 left

💡 Hint

Common Mistakes

Using %c which prints only one character

Using %d or %f which are for numbers

The %s format specifier is used to print strings in C.

3fill in blank

hard

Fix the error in copying a string using strcpy.

DSA C

char dest[20];
strcpy(dest, [1]);

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

Asource

Bstrcpy

C"source"

Ddest

Attempts:

3 left

💡 Hint

Common Mistakes

Using destination as source

Using string literal instead of variable

The second argument to strcpy should be the source string variable, not the destination or a string literal.

4fill in blank

hard

Fill both blanks to create a loop that prints each character of a string until the null character.

DSA C

int i = 0;
while (str[i] [1] '\0') {
    printf("%c", str[i]);
    i[2];
}

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

A!=

B==

C++

D--

Attempts:

3 left

💡 Hint

Common Mistakes

Using == instead of !=

Decrementing i instead of incrementing

The loop continues while the character is not the null character '\0'. The index i is incremented each time.

5fill in blank

hard

Fill all three blanks to create a function that returns the length of a string.

DSA C

int string_length(char *str) {
    int [1] = 0;
    while (str[[2]] != [3]) {
        length++;
    }
    return length;
}

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

Alength

Bi

C'\0'

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong variable names

Not checking for '\0' correctly

We use a variable length initialized to 0, check characters until the null character '\0', and increment length.

Practice

1. In C, how is a string typically represented in memory?

easy

A. As a built-in string object with methods

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

C. As a linked list of characters

D. As a fixed-length integer array

Solution

1. Step 1: Understand C string representation

   In C, strings are arrays of characters stored in contiguous memory locations.

2. Step 2: Identify string termination

   C strings end with a special null character '\0' to mark the end of the string.

3. Final Answer:

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

4. Quick Check:

   C strings = char array + '\0' [OK]

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

Common Mistakes:

• Thinking strings are objects with methods like in other languages
• Assuming strings are linked lists in C
• Ignoring the null terminator character

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 = 'hello';

C. string str = "hello";

D. char str = "hello";

Solution

1. Step 1: Check declaration syntax for string literals in C

   In C, strings are arrays of characters, so we declare with char array syntax.

2. Step 2: Validate initialization

   char str[] = "hello"; creates an array with characters 'h','e','l','l','o','\0'.

3. Final Answer:

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

4. Quick Check:

   char array with double quotes = correct string declaration [OK]

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

Common Mistakes:

• Using single quotes for strings instead of double quotes
• Trying to use 'string' type which doesn't exist in C
• Declaring char as a single character instead of array

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

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

medium

A. \0 (null character, no visible output)

B. c

C. a

D. Runtime error

Solution

1. Step 1: Understand string indexing in C

   String "abc" is stored as ['a','b','c','\0'] with indices 0 to 3.

2. Step 2: Identify character at index 3

   str[3] is the null terminator '\0', which prints no visible character.

3. Final Answer:

   \0 (null character, no visible output) -> Option A

4. Quick Check:

   Index 3 = '\0' null terminator [OK]

Hint: Remember last char in C string is '\0' at length index [OK]

Common Mistakes:

• Expecting str[3] to print 'c' instead of null
• Assuming runtime error due to out-of-bounds access
• Confusing zero-based indexing

4. Identify the error in this C code snippet that attempts to copy a string:

char src[] = "hello";
char dest[5];
strcpy(dest, src);
printf("%s", dest);

medium

A. strcpy function is not declared

B. printf format specifier should be %c instead of %s

C. src should be a pointer, not an array

D. dest array size is too small to hold the string and null terminator

Solution

1. Step 1: Check size of destination array

   "hello" has 5 characters plus '\0' terminator, total 6 chars needed.

2. Step 2: Compare with dest array size

   dest is declared with size 5, which is too small to hold full string plus '\0'.

3. Final Answer:

   dest array size is too small to hold the string and null terminator -> Option D

4. Quick Check:

   Destination must fit string + '\0' [OK]

Hint: Always allocate space for string length + 1 for '\0' [OK]

Common Mistakes:

• Ignoring null terminator space in array size
• Assuming strcpy needs special declaration
• Confusing %c and %s in printf

5. Given the following C code, what will be the output?

char str1[] = "test";
char *str2 = "test";
str1[0] = 'b';
// str2[0] = 'b'; // commented out
printf("%s %s", str1, str2);

Why is the line modifying str2 commented out?

hard

A. Because str2 is an array and arrays cannot be modified

B. Because str2 is a constant pointer and cannot be changed

C. Because string literals pointed by str2 are stored in read-only memory and cannot be modified

D. Because str2 points to a copy of str1 which is immutable

Solution

1. Step 1: Understand difference between array and pointer to string literal

   str1 is a modifiable array copy of "test"; str2 points to a string literal in read-only memory.

2. Step 2: Explain why modifying str2 causes error

   Modifying string literals is undefined behavior and often causes runtime error, so line is commented out.

3. Final Answer:

   Because string literals pointed by str2 are stored in read-only memory and cannot be modified -> Option C

4. Quick Check:

   String literals are read-only, arrays are modifiable [OK]

Hint: String literals are read-only; arrays can be changed [OK]

Common Mistakes:

• Thinking str2 is a constant pointer (it's pointer to const data)
• Assuming arrays cannot be modified
• Confusing pointer and array behavior