String Traversal and Character Access in DSA C - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of character traversal with pointer increment

What is the output of this C code snippet that traverses a string using a pointer?

DSA C

char *str = "hello";
char *p = str;
while (*p) {
    printf("%c-", *p);
    p++;
}
printf("null");

A h e l l o null

B h-e-l-l-o-

C hello-null

D h-e-l-l-o-null

💡 Hint

Remember that the loop prints each character followed by a dash until it reaches the null character.

The loop prints each character followed by a dash. After the loop ends, it prints 'null'. So the output is characters separated by dashes ending with 'null'.

❓ Predict Output

intermediate

Value of character after pointer arithmetic

Given the code below, what character is printed by the printf statement?

DSA C

char str[] = "world";
char *p = str + 2;
printf("%c", *p);

A l

B r

C o

D d

💡 Hint

Pointer p points to the third character in the string (index 2).

The string "world" has characters w(0), o(1), r(2), l(3), d(4). Pointer p points to index 2, which is 'r'.

🔧 Debug

advanced

Identify the error in string traversal loop

What error does this code produce when compiled and run?

DSA C

char *s = "test";
int i = 0;
while (s[i] != '\0') {
    printf("%c", s[0]);
    i--;
}

A Infinite loop printing 't'

B Compilation error

C Prints 'test' once and stops

D Segmentation fault

💡 Hint

Check how the index variable changes inside the loop.

The index i is decremented inside the loop, so it goes negative and s[i] never reaches '\0'. This causes an infinite loop printing the first character repeatedly.

❓ Predict Output

advanced

Output of nested loops accessing string characters

What is the output of this code snippet?

DSA C

char str[] = "abc";
for (int i = 0; i < 3; i++) {
    for (int j = 0; j <= i; j++) {
        printf("%c", str[j]);
    }
    printf("-");
}

A a-ab-abc-

B abc-abc-abc-

C a-aa-aaa-

D a-bc-abc-

💡 Hint

Inner loop prints characters from start up to current outer loop index.

For i=0, prints 'a'; i=1 prints 'ab'; i=2 prints 'abc'. Each followed by a dash.

🧠 Conceptual

expert

Number of characters printed by pointer loop

How many characters are printed by this code snippet?

DSA C

char *p = "data";
int count = 0;
while (*p++) {
    count++;
}
printf("%d", count);

A 3

B 5

C 4

D 0

💡 Hint

The loop counts characters until the null terminator is reached.

The string "data" has 4 characters before the null terminator. The loop increments count for each character until it reaches '\0'.

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