String Traversal and Character Access in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to print each character of the string using a loop.

DSA C

char str[] = "hello";
for (int i = 0; i < [1]; i++) {
    printf("%c\n", str[i]);
}

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

Astr.length

B5

Cstrlen(str)

Dsizeof(str)

Attempts:

3 left

💡 Hint

Common Mistakes

Using sizeof(str) which includes the null terminator and size of array.

Using a fixed number without considering string length.

strlen(str) returns the length of the string excluding the null character, so the loop prints all characters.

2fill in blank

medium

Complete the code to access the last character of the string.

DSA C

char str[] = "world";
char last_char = str[[1]];
printf("%c\n", last_char);

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

Astrlen(str) - 1

Bsizeof(str) - 1

Cstrlen(str)

Dsizeof(str)

Attempts:

3 left

💡 Hint

Common Mistakes

Using strlen(str) as index which is out of bounds.

Using sizeof(str) which includes null terminator.

The last character is at index length - 1 because indexing starts at 0.

3fill in blank

hard

Fix the error in the code to correctly print characters until the null terminator.

DSA C

char str[] = "test";
int i = 0;
while (str[[1]] != '\0') {
    printf("%c\n", str[i]);
    i++;
}

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

Astrlen(str)

B0

C1

Di

Attempts:

3 left

💡 Hint

Common Mistakes

Using a fixed index like 0 or 1 in the condition causes infinite loop or wrong output.

Using strlen(str) inside the condition without updating index.

The loop should check str[i] to stop at the null terminator correctly.

4fill in blank

hard

Fill both blanks to create a loop that prints characters only if they are vowels.

DSA C

char str[] = "example";
for (int i = 0; i < [1]; i++) {
    if (str[i] == [2]) {
        printf("%c\n", str[i]);
    }
}

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

Astrlen(str)

B'a'

C'e'

D'x'

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong loop limit causing out of bounds.

Checking for a consonant instead of a vowel.

Loop runs through the string length and prints characters only if they are 'e' (a vowel).

5fill in blank

hard

Fill all three blanks to create a loop that counts how many times 'l' appears in the string.

DSA C

char str[] = "hello world";
int count = 0;
for (int i = 0; i < [1]; i++) {
    if (str[i] == [2]) {
        [3];
    }
}
printf("%d\n", count);

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

Astrlen(str)

B'l'

Ccount++

Di++

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong character in condition.

Not incrementing count inside if block.

Loop runs through the string length, checks if character is 'l', and increments count.

Practice

1. What does the expression str[0] represent when str is a string in C?

easy

A. The first character of the string

B. The length of the string

C. The last character of the string

D. A pointer to the string

Solution

1. Step 1: Understand string indexing in C

   In C, strings are arrays of characters. Indexing starts at 0, so str[0] accesses the first character.

2. Step 2: Identify what str[0] returns

   It returns the character stored at the first position of the string, not the length or pointer.

3. Final Answer:

   The first character of the string -> Option A

4. Quick Check:

   str[0] = first character [OK]

Hint: Remember: string indexing starts at zero in C [OK]

Common Mistakes:

• Confusing character access with string length
• Thinking indexing starts at 1
• Mixing pointer and character access

2. Which of the following is the correct way to traverse a string str of length n in C using a for loop?

easy

A. for (int i = 1; i < n; i++)

B. for (int i = 0; i <= n; i++)

C. for (int i = 0; i < n; i++)

D. for (int i = 1; i <= n; i++)

Solution

1. Step 1: Understand string indexing and loop bounds

   String indices in C start at 0 and go up to n-1 for length n.

2. Step 2: Choose loop condition to cover all characters

   The loop should run from 0 to n-1, so i < n is correct.

3. Final Answer:

   for (int i = 0; i < n; i++) -> Option C

4. Quick Check:

   Loop from 0 to n-1 = for (i=0; i<n; i++) [OK]

Hint: Loop from 0 to less than length for strings [OK]

Common Mistakes:

• Using <= n causes out-of-bounds access
• Starting loop at 1 misses first character
• Confusing string length with last index

3. What is the output of the following C code?

char str[] = "hello";
for (int i = 0; i < 5; i++) {
    printf("%c-", str[i]);
}

medium

A. hello-

B. h-e-l-l-o-

C. h e l l o

D. h-e-l-l-o

Solution

1. Step 1: Analyze the loop and print statement

   The loop runs from i=0 to 4, printing each character followed by a dash.

2. Step 2: Trace each iteration output

   Characters printed: 'h-', 'e-', 'l-', 'l-', 'o-'. Combined output is "h-e-l-l-o-".

3. Final Answer:

   h-e-l-l-o- -> Option B

4. Quick Check:

   Each char + '-' printed = h-e-l-l-o- [OK]

Hint: Look at printf format and loop range carefully [OK]

Common Mistakes:

• Ignoring the dash after each character
• Printing the whole string once instead of chars
• Confusing spaces with dashes

4. Identify the error in the following code snippet that tries to print each character of a string str.

char str[] = "world";
for (int i = 1; i <= 5; i++) {
    printf("%c", str[i]);
}

medium

A. Both A and B are errors

B. Loop condition should be i < 5 instead of <= 5

C. No error, code works fine

D. Loop starts at 1 and misses the first character

Solution

1. Step 1: Check loop start index

   Loop starts at 1, so it skips str[0] which is the first character.

2. Step 2: Check loop end condition

   Loop runs while i <= 5, but valid indices are 0 to 4. Accessing str[5] causes out-of-bounds error.

3. Final Answer:

   Both A and B are errors -> Option A

4. Quick Check:

   Start at 0 and use i < length to avoid errors [OK]

Hint: Loop from 0 to less than length to avoid errors [OK]

Common Mistakes:

• Starting loop at 1 and missing first char
• Using <= length causing out-of-bounds access
• Ignoring array indexing rules

5. Given a string str of unknown length, which of the following code snippets correctly prints all characters until the null terminator in C?

hard

A. for (int i = 0; str[i] != '\0'; i++) { printf("%c", str[i]); }

B. for (int i = 0; i < strlen(str); i++) { printf("%c", str[i]); }

C. while (*str != '\0') { printf("%c", *str); str++; }

D. All of the above

Solution

1. Step 1: Analyze for (int i = 0; str[i] != '\0'; i++) { printf("%c", str[i]); }

   Loops until character is not null terminator, printing each character. Correct method.

2. Step 2: Analyze for (int i = 0; i < strlen(str); i++) { printf("%c", str[i]); } and C

   for (int i = 0; i < strlen(str); i++) { printf("%c", str[i]); } uses strlen to get length, then loops printing characters. while (*str != '\0') { printf("%c", *str); str++; } uses pointer traversal until null terminator. Both are correct.

3. Step 3: Combine results

   All three methods correctly print the string characters until the null terminator.

4. Final Answer:

   All of the above -> Option D

5. Quick Check:

   All methods stop at '\0' and print chars [OK]

Hint: Any loop stopping at '\0' prints full string safely [OK]

Common Mistakes:

• Not checking for null terminator causing overflow
• Using wrong loop conditions
• Modifying string pointer incorrectly