Lab 9 – Active Strings & Characters

PSUC Lab 9 — Null-Terminated Arrays

Active Logic: Strings & Null Terminator `'\0'`

Master C string manipulation: ASCII offset conversion, last character occurrence search, string.h library functions (strlen, strcpy, strcat), word count, string reversal, and palindrome checks.

C String Memory Rules

Concept	Rule / Formula	Key Trap / Pitfall
Null Terminator	`str[n] == '\0'` (ASCII 0)	Omitting `'\0'` causes memory overrun!
Case Conversion	`'a' - 'A' = 32`	`c - 32` converts lowercase to uppercase
Word Count	Count spaces + 1	Must ignore leading/consecutive spaces!
Palindrome	`str[i] == str[len - 1 - i]`	Compare 2 pointers moving inward!

🎯 Learning Philosophy: Active Logic Building

Test your string logic across 3 Interactive Learning Modes:

🧠 Tab 1 (Logic & Hints): String memory & null terminator rules.
🧩 Tab 2 (Logic Puzzle): Fill in missing ASCII offsets/indices & test live text tools!
💻 Tab 3 (C Solution): Reference verified C code solutions.

Solve string logic puzzles before inspecting C source code!

Exercise 1 of 5

Convert Lower-case Letters to Upper-case

Converts all lower-case characters in a string to upper-case using ASCII offset calculation (c - 32).

ASCII Table Offset

'a' = ASCII 97 | 'A' = ASCII 65
Offset = 97 - 65 = 32
Formula: if (c >= 'a' && c <= 'z') c = c - 32;

Interactive Logic Hints

💡 Hint 1: What does toupper() in do internally? ▼

It checks if character is lowercase and subtracts 32 automatically!

🧩 Complete Lowercase to Uppercase Loop

while (str[i] != ) {
  if (str[i] >= 'a' && str[i] <= 'z') {
    str[i] = str[i] ;
  }
  i++;
}

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🔤 Uppercase Converter Simulator

Input Text:

Converted text...

lower_to_upper.cReference C Code

#include <stdio.h>

int main() {
    char str[] = "welcome to psuc visual lab";
    int i = 0;

    while (str[i] != '\0') {
        if (str[i] >= 'a' && str[i] <= 'z') {
            str[i] = str[i] - 32;
        }
        i++;
    }

    printf("Uppercase: %s\n", str);
    return 0;
}

Exercise 2 of 5

Find Last Occurrence of a Character

Traverses string to find the last index position where character key appears.

Reverse Search Optimization

Method 1: Forward scan & overwrite lastIdx = i when match found.
Method 2: Backward scan from strlen(str) - 1 down to 0 & break on first match!

Interactive Logic Hints

💡 Hint 1: Which method is faster for finding LAST occurrence? ▼

Backward scan is faster! It finds the last occurrence immediately and breaks!

🧩 Complete Backward Character Search Loop

int len = strlen(str), lastIdx = -1;
for (i = len - 1; i 0; i--) {
  if (str[i] key) {
    lastIdx = i; ;
  }
}

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🔍 Last Occurrence Search Visualizer

String:

Character Key:

Search status...

last_char_search.cReference C Code

#include <stdio.h>
#include <string.h>

int main() {
    char str[] = "programming in c language";
    char key = 'g';
    int i, len = strlen(str), lastIdx = -1;

    for (i = len - 1; i >= 0; i--) {
        if (str[i] == key) {
            lastIdx = i;
            break;
        }
    }

    if (lastIdx != -1) printf("Last occurrence of '%c' is at index %d\n", key, lastIdx);
    else               printf("Character '%c' NOT found.\n", key);
    return 0;
}

Exercise 3 of 5

Count Number of Words in a Sentence

Counts total words by scanning spaces and non-space word transitions.

Word Counting State Machine

Track state: inWord = false
If char is non-space AND !inWord → words++; inWord = true;
If char is space → inWord = false;

Interactive Logic Hints

💡 Hint 1: Why not just count spaces + 1? ▼

Counting spaces fails if user types multiple consecutive spaces ("hello world")! State tracking handles extra spaces correctly!

🧩 Complete Word Count Logic

int words = 0; bool inWord = false;
while (str[i] != '\0') {
  if (str[i] != ' ' && str[i] != '\t' && str[i] != '\n') {
    if (!inWord) { words; inWord = ; }
  } else {
    inWord = false;
  }
  i++;
}

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📝 Word Counter Visualizer

Sentence:

Word count...

word_count.cReference C Code

#include <stdio.h>
#include <stdbool.h>

int main() {
    char str[] = "C programming is essential for computer engineering";
    int i = 0, words = 0;
    bool inWord = false;

    while (str[i] != '\0') {
        if (str[i] != ' ' && str[i] != '\t' && str[i] != '\n') {
            if (!inWord) { words++; inWord = true; }
        } else {
            inWord = false;
        }
        i++;
    }

    printf("Total words = %d\n", words);
    return 0;
}

Exercise 4 of 5

String Reversal & Palindrome Check

Reverses string using two-pointer swap and checks if str[i] == str[len - 1 - i] for all indices.

Two-Pointer In-Place Swap

Pointers: left = 0, right = len - 1
Swap: temp = str[left]; str[left] = str[right]; str[right] = temp;
Increment/Decrement: left++; right--;

Interactive Logic Hints

💡 Hint 1: What are famous string palindromes? ▼

"madam", "racecar", "level", "radar"!

🧩 Complete Two-Pointer Reversal Loop

int left = 0, right = len - 1;
while (left right) {
  char temp = str[left];
  str[left] = str[];
  str[right] = temp;
  left++; right--;
}

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🔄 String Reversal & Palindrome Engine

Enter Word/String:

Reversal status...

string_reverse.cReference C Code

#include <stdio.h>
#include <string.h>
#include <stdbool.h>

int main() {
    char str[] = "racecar";
    int i, len = strlen(str);
    bool isPal = true;

    for (i = 0; i < len / 2; i++) {
        if (str[i] != str[len - 1 - i]) {
            isPal = false;
            break;
        }
    }

    printf("String '%s' IS %sa Palindrome!\n", str, isPal ? "" : "NOT ");
    return 0;
}

Exercise 5 of 5

Manual Strlen & Substring Extraction

Computes string length without library functions and extracts sub-string from index start for length characters.

Manual Strlen & Substring Rules

Manual Strlen: while (str[len] != '\0') len++;
Substring Extraction: Copy sub[i] = str[start + i] for i = 0 to subLen - 1, then append sub[subLen] = '\0'!

Interactive Logic Hints

💡 Hint 1: What happens if you forget sub[subLen] = '\0'? ▼

The extracted substring will print garbage characters! Always null-terminate new strings!

🧩 Complete Substring Extraction Loop

for (i = 0; i < subLen; i++) {
sub[i] = str[start i];
}
sub[subLen] = ;

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✂️ Substring Extractor Tool

Start Position:

Substring Length:

Extracted Substring...

substring_manual.cReference C Code

#include <stdio.h>

int main() {
    char str[] = "Manipal University Jaipur";
    char sub[50];
    int start = 8, length = 10, i;

    for (i = 0; i < length; i++) {
        sub[i] = str[start + i];
    }
    sub[length] = '\0';

    printf("Extracted Substring: '%s'\n", sub);
    return 0;
}

Knowledge Check

Test Your String Logic Mastery

Question 1 of 5

Loading question...

Active Logic: Strings & Null Terminator '\0'

C String Memory Rules

🎯 Learning Philosophy: Active Logic Building

Convert Lower-case Letters to Upper-case

ASCII Table Offset

Interactive Logic Hints

🧩 Complete Lowercase to Uppercase Loop

🔤 Uppercase Converter Simulator

Find Last Occurrence of a Character

Reverse Search Optimization

Interactive Logic Hints

🧩 Complete Backward Character Search Loop

🔍 Last Occurrence Search Visualizer

Count Number of Words in a Sentence

Word Counting State Machine

Interactive Logic Hints

🧩 Complete Word Count Logic

📝 Word Counter Visualizer

String Reversal & Palindrome Check

Two-Pointer In-Place Swap

Interactive Logic Hints

🧩 Complete Two-Pointer Reversal Loop

🔄 String Reversal & Palindrome Engine

Manual Strlen & Substring Extraction

Manual Strlen & Substring Rules

Interactive Logic Hints

🧩 Complete Substring Extraction Loop

✂️ Substring Extractor Tool

Test Your String Logic Mastery

Active Logic: Strings & Null Terminator `'\0'`