String Traversal and Character Access in DSA C

Mental Model

A string is a sequence of characters stored one after another. Traversing means visiting each character one by one to read or process it.

Analogy: Imagine reading a sentence letter by letter from left to right, like reading a book page from start to end.

S -> t -> r -> i -> n -> g -> \0
↑
(start)

Dry Run Walkthrough

Input: string: "cat"

Goal: Print each character of the string one by one in order

Step 1: Access first character 'c'

c -> a -> t -> \0
↑

Why: Start reading from the first character

Step 2: Access second character 'a'

c -> a -> t -> \0
  ↑

Why: Move to next character to continue reading

Step 3: Access third character 't'

c -> a -> t -> \0
    ↑

Why: Move to next character to continue reading

Step 4: Detect null character '\0' to stop

c -> a -> t -> \0
      ↑

Why: Null character marks end of string, stop traversal

Result:

c\na\nt\n

Annotated Code

DSA C

#include <stdio.h>

void traverseString(const char *str) {
    int i = 0;
    while (str[i] != '\0') { // check for end of string
        printf("%c\n", str[i]); // print current character
        i++; // move to next character
    }
}

int main() {
    const char *myString = "cat";
    traverseString(myString);
    return 0;
}

while (str[i] != '\0') {

check if current character is not end marker to continue traversal

printf("%c\n", str[i]);

print the current character

i++;

advance index to next character

Output

c

a

t

Complexity Analysis

Time: O(n) because we visit each character once until the null terminator

Space: O(1) because we use a fixed number of variables regardless of string size

vs Alternative: Compared to recursive traversal, this iterative approach uses less memory and is simpler

Edge Cases

empty string ""

The loop does not run and nothing is printed

DSA C

while (str[i] != '\0') {

string with one character "a"

Prints the single character then stops

DSA C

while (str[i] != '\0') {

When to Use This Pattern

When you need to process or examine each character in a string one by one, use string traversal with a loop until the null character.

Common Mistakes

Mistake: Forgetting to check for the null character '\0' and causing out-of-bounds access
Fix: Always check str[i] != '\0' before accessing str[i]

Mistake: Using incorrect loop condition like i < length without knowing string length
Fix: Use null character check to safely traverse strings of unknown length

Summary

It visits each character in a string one by one until the end.

Use it when you want to read or process every character in a string.

The null character '\0' marks the end and stops traversal.

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