II.15 Structures, Unions, and Bit Manipulation

⬆️ P-I.II. Programming Fundamentals · Question Bank · Chapter II (2016–2026)

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Remember: Struct stores together; union shares; masks select bits.

Start here — what you should be able to do

• Design records and pass them to functions.
• Explain union storage and active members accurately.
• Calculate bitwise results and use valid masks.

Reading order: foundation checks → full notes and examples → exam traps → self-test without notes → past-paper answers. All sections below remain part of the topic; code examples are kept in full.

Foundation checks

Bit-mask toolkit

Use an unsigned value x and an unsigned mask 1u << k, with k within the type's bit width.

Memory: ordinary struct members have separate storage, with possible padding. A union has shared storage sufficient for its largest member plus alignment requirements. Active member: do not assume reading arbitrary inactive union members is portable C++. Enum: unassigned enumerators start at zero and then increment; an explicit value changes the subsequent sequence.

1. Structures (struct)

• Groups different data types under one name; each member gets its own separate memory, so the struct's total size is (at least) the sum of its members' sizes

struct Student {
    char name[20];
    int age;
    float gpa;
};
struct Student s1 = {"Ali", 20, 3.8};
printf("%s", s1.name);       // dot operator to access members

• Members are accessed with the dot operator (.) on a struct variable, or the arrow operator (->) via a pointer to a struct: ptr->name
• Nested structures: a struct can contain another struct as a member
• Structures can be passed to functions by value (a full copy) or by reference/pointer (to avoid copying and/or allow modification)

2. Unions

• Union: members share storage; size is sufficient for the largest member and alignment requirements, not the sum.

union Data {
    int i;
    float f;
    char c;
};
union Data d;
d.i = 10;     // writing to i overwrites the same memory f and c would use

• Useful when only one of several possible types is needed at a time (memory-saving), e.g. representing a value that could be an int OR a float depending on context
• Active member: ordinary C++ union use has one active member; arbitrary inactive-member reads are not portable type-punning.

3. Struct vs Union

4. Enumerations (enum)

• A user-defined type consisting of a set of named integer constants, improving readability over "magic numbers"

enum Day {SUN, MON, TUE, WED, THU, FRI, SAT};   // SUN=0, MON=1, ... by default
enum Day today = WED;   // today holds the integer value 3

5. Bit Manipulation Operators

• Example: 6 & 3 → 0110 & 0011 = 0010 = 2; 6 | 3 → 0110 | 0011 = 0111 = 7; 6 ^ 3 → 0101 = 5
• 5 << 1 = 10 (multiply by 2); 5 >> 1 = 2 (divide by 2, truncated)

6. Bit Masking / Flags Technique

• A common low-level pattern for storing multiple boolean flags compactly in one integer:

#define READ  1   // 001
#define WRITE 2   // 010
#define EXEC  4   // 100
int perms = READ | WRITE;      // set flags: 011
if (perms & WRITE) { ... }      // check a flag: is WRITE set?
perms = perms & ~WRITE;         // clear a flag: unset WRITE

• Used in permission systems, hardware registers, and status flags where memory efficiency matters

7. 💻 Code Examples: C & C++

Struct, union, enum, and a bit-flag permissions example, all in one program:

C:

#include <stdio.h>
struct Student { char name[20]; int age; float gpa; };
union Data { int i; float f; };
enum Day { SUN, MON, TUE, WED, THU, FRI, SAT };

#define READ  1   // 001
#define WRITE 2   // 010

int main(void) {
    struct Student s1 = {"Ali", 20, 3.8};
    printf("%s is %d, GPA %.1f\n", s1.name, s1.age, s1.gpa);  // dot operator

    union Data d;
    d.i = 10;
- **Size example:** the union's size accommodates its largest member and alignment; int/float are often four bytes, but check the actual implementation.

    enum Day today = WED;
    printf("today = %d\n", today);            // 3

    int perms = READ | WRITE;                   // set flags: 011
    if (perms & WRITE) printf("write allowed\n");  // check a flag
    perms = perms & ~WRITE;                       // clear a flag
    return 0;
}

C++:

#include <iostream>
using namespace std;
struct Student { string name; int age; float gpa; };  // C++ allows std::string as a member
union Data { int i; float f; };
enum Day { SUN, MON, TUE, WED, THU, FRI, SAT };

const int READ  = 1;  // 001
const int WRITE = 2;  // 010

int main() {
    Student s1{"Ali", 20, 3.8f};               // C++11 brace initialization
    cout << s1.name << " is " << s1.age << ", GPA " << s1.gpa << endl;

    Data d;
    d.i = 10;
    cout << "d.i=" << d.i << " (sizeof union = " << sizeof(d) << ")" << endl;

    Day today = WED;
    cout << "today = " << today << endl;         // 3

    int perms = READ | WRITE;
    if (perms & WRITE) cout << "write allowed" << endl;
    perms = perms & ~WRITE;
    return 0;
}

• Size example: the union's size accommodates its largest member and alignment; int/float are often four bytes, but check the actual implementation.

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8. MCQ Traps

• In a struct, each member has separate memory; in a union, all members share the same memory
• Union size: at least enough for the largest member, with alignment/padding as required.
• Accessing a struct member via a pointer requires the arrow operator (->), not the dot operator
• enum values default to 0, 1, 2, ... in declaration order unless explicitly assigned otherwise
• Left shift (<<) multiplies by 2 per shift; right shift (>>) divides by 2 per shift — but only under the appropriate unsigned/range and valid-shift-count conditions

9. Self-Test Questions — with answers

Q1. Why is union storage based on the largest member?

Answer

• Members overlap the same storage.
• It must be sufficient for the largest member and satisfy alignment requirements; it is not the sum of member sizes.
• Ordinary C++ use has one active member at a time.

Q2. Which operator accesses a member through a pointer?

Answer

• ptr->member, equivalent to (*ptr).member.
• Use object.member when you have the struct object itself.

Q3. Compute 12 AND 10, OR 10 and XOR 10.

Answer

• 12 = 1100; 10 = 1010.
• AND: 1000 = 8.
• OR: 1110 = 14.
• XOR: 0110 = 6.

Q4. What values do JAN, FEB and MAR have by default?

Answer

• For enum Month { JAN, FEB, MAR };: 0, 1 and 2.
• If JAN is explicitly 1, the following unassigned values become 2 and 3.

Q5. How do OR and AND set and test a flag?

Answer

• Set: flags |= mask.
• Test: (flags & mask) != 0.
• Clear: flags &= ~mask.
• Use an unsigned mask and a valid bit position.

1. Why does a union's size equal its largest member's size rather than the sum of all members?
2. What operator is used to access a struct member through a pointer, and why?
3. Compute 12 & 10, 12 | 10, and 12 ^ 10 in binary.
4. What integer values do the enum constants {JAN, FEB, MAR} take by default?
5. Describe how bitwise OR and AND can be used to set and check a flag bit, respectively.

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10. 📝 Exam Practice: Binary Logic Gate Simulator

Q: Write a C/C++ program which implements binary logical 'AND', 'OR', and 'XOR' gates. The program takes binary numbers and a desired logical gate, and outputs the result of the gate.

Solution:

#include <iostream>
#include <string>
using namespace std;

int main()
{
    string bin1, bin2, gate;

    cout << "Enter first binary number: ";
    cin >> bin1;

    cout << "Enter second binary number: ";
    cin >> bin2;

    cout << "Enter desired gate (AND / OR / XOR): ";
    cin >> gate;

    // Pad the shorter string with leading zeros so both are equal length
    int maxLen = (bin1.length() > bin2.length()) ? bin1.length() : bin2.length();
    while (bin1.length() < maxLen) bin1 = "0" + bin1;
    while (bin2.length() < maxLen) bin2 = "0" + bin2;

    string result = "";
    for (int i = 0; i < maxLen; i++)
    {
        int bit1 = bin1[i] - '0';   // convert char '0'/'1' to int 0/1
        int bit2 = bin2[i] - '0';
        int resBit;

        if (gate == "AND" || gate == "and")      resBit = bit1 & bit2;
        else if (gate == "OR" || gate == "or")   resBit = bit1 | bit2;
        else if (gate == "XOR" || gate == "xor") resBit = bit1 ^ bit2;
        else { cout << "Invalid gate entered."; return 0; }

        result += (resBit == 1) ? '1' : '0';
    }

    cout << "\nResult of " << gate << " operation: " << result << endl;
    return 0;
}

Sample run:

Enter first binary number: 1010
Enter second binary number: 1100
Enter desired gate (AND / OR / XOR): AND

Result of AND operation: 1000

Column-method trace (matches the textbook's "evaluate by column" approach):

Key reasoning points:

• Binary numbers are treated as strings, not integers — each character is processed individually with &, |, ^ applied to single 0/1 ints, mirroring the textbook's column-by-column method
• Shorter input is zero-padded on the left first, so both operands align at the same bit positions
• Matches the chapter's summary rules exactly: AND needs both bits 1, OR needs at least one bit 1, XOR needs an odd count of 1 bits (here just "the two bits differ")

11. 📜 Past-Paper Questions & MCQs (CSS 2016–2026)

2016 · Paper-I · Q4(a) · 6 marks

Recall cue: Define the record fields; input the complete record; test age boundaries; display the category.

Q. Define a structure for a Person (name, age, address). Input one record and display "Young" if age < 20, "Old" if age > 50, otherwise "Middle".

Answer — key points

struct Person { char name[30]; int age; char address[50]; };
int main() {
    Person p;
    cout << "Enter Name = ";    cin.getline(p.name, 30);
    cout << "Enter Age = ";     cin >> p.age;  cin.ignore();
    cout << "Enter Address = "; cin.getline(p.address, 50);
    if (p.age < 20)      cout << "Person is YOUNG";
    else if (p.age > 50) cout << "Person is OLD";
    else                 cout << "Person is MIDDLE";
}

• The paper leaves ages 20 and 50 undefined, so treat them as Middle.

2017 · Paper-I · Q3(c) · 6 marks

Recall cue: Store code, name and price together; find the code; multiply price by quantity; accumulate the bill.

Q. Write a supermarket checkout program. The database of items (name, code, price) is a structure. The cashier enters codes and quantities, and the program prints the total.

Answer — key points

struct Item { int code; char name[20]; float price; };
Item db[3] = { {1,"Rice",500}, {2,"Sugar",150}, {3,"Oil",400} };
float total = 0; char more;
do {
    int code, qty;
    cout << "Enter product code: ";  cin >> code;
    cout << "Enter quantity: ";      cin >> qty;
    for (int i = 0; i < 3; i++)
        if (db[i].code == code) total += db[i].price * qty;
    cout << "Do you have more products to add (Y/N): "; cin >> more;
} while (more == 'Y' || more == 'y');
cout << "The total amount is " << total;

• To search by name, compare with strcmp(db[i].name, input) == 0.

2019 · Paper-I · Q5(b) · 5 marks and 2023 · Paper-I · Q3(c) · 6 marks

Recall cue: Define bitwise operations; demonstrate masks; use unsigned values and valid shift counts.

Q. What is bit twiddling? Give a brief description.

Answer — key points

• Bit twiddling (bit manipulation) means working directly on the individual bits of data using bitwise operators &, |, ^, ~, <<, >>.
• Why: it is fast and memory-efficient, so it is used in device drivers, graphics, cryptography, compression and flags.
• Examples:
  • odd or even: n & 1
  • multiply by 2: n << 1
  • set bit k: n | (1 << k)
  • clear bit k: n & ~(1 << k)
  • toggle bit k: n ^ (1 << k)
  • power of two test: n != 0 && (n & (n - 1)) == 0
  • swap without a temporary: a ^= b; b ^= a; a ^= b;

2022 · Paper-I · Q4(b) · 8 marks

Recall cue: Write both numbers in binary; OR each bit; convert 11101 back to 29.

Q. Write the output of the program with a = 12, b = 25 that prints a | b.

Answer — key points

• 12 = 01100, 25 = 11001, so 12 | 25 = 11101 = 29.
• Output:

a = 12
b = 25
a | b = 29

2025 · Paper-I · Q4(c) · 8 marks

Recall cue: Represent real and imaginary parts; add corresponding fields; format negative imaginary parts correctly.

Q. Demonstrate two complex numbers using a structure. Use a function to add them and display the result as (a + bi).

Answer — key points

struct Complex { float re, im; };
Complex add(Complex x, Complex y) { return { x.re + y.re, x.im + y.im }; }
void show(Complex c) {
    cout << "(" << c.re << (c.im < 0 ? " - " : " + ") << (c.im < 0 ? -c.im : c.im) << "i)";
}
int main() {
    Complex a, b;
    cout << "First (real imag): ";  cin >> a.re >> a.im;
    cout << "Second (real imag): "; cin >> b.re >> b.im;
    show(add(a, b));
}

2026 · Paper-I · Q3(a) · 6 marks

Recall cue: Explain bit manipulation; show checking and setting/clearing bits; avoid blanket performance claims.

Q. What is bit manipulation in programming, and why is it used? Give two examples.

Answer — key points

• Bit manipulation uses bitwise operators (&, |, ^, ~, <<, >>) to read or change individual bits.
• Why used: speed, low memory use, and direct control of flags, masks, hardware registers, compression and encryption.
• Example 1: checking odd or even with n & 1 (1 means odd).
• Example 2: setting and clearing a bit with flags |= (1 << 3) and flags &= ~(1 << 3).

MCQs

2026 P-I · Q3. A programmer wants to group different data types. What should he or she use?

• (A) Structures ✅
• (B) Arrays
• (C) Unions
• (D) None of these
• Answer: (A) Structures.
• Why: a structure holds members of different types together, each with its own memory (for example an int, a float and a char array). An array holds elements of one type only. A union also mixes types, but all members share the same memory, so only one value is valid at a time.

Technical accuracy notes

• For unsigned n, a power-of-two test must exclude zero: n != 0 && (n & (n - 1)) == 0.
• Use unsigned masks such as 1u << k, with k less than the type's bit width. Bitwise operations are not automatically faster than arithmetic after optimisation.
• A union is large enough for its largest member and obeys alignment requirements. Its members share storage, but arbitrary access through an inactive member is not portable C++ type-punning. C++ working draft.
• Prefer a temporary variable or an appropriate swap function over XOR swapping, particularly when two expressions may alias.

Recall check and coverage

• Explain the three learning outcomes at the top without looking.
• Reproduce one example or trace and test a boundary case.
• Answer each self-test, then compare with the key.
• Attempt each recorded past-paper task on this page.
• Recorded written references: 2016 · Paper-I · Q4(a) · 6 marks; 2017 · Paper-I · Q3(c) · 6 marks; 2019 · Paper-I · Q5(b) · 5 marks and 2023 · Paper-I · Q3(c) · 6 marks; 2022 · Paper-I · Q4(b) · 8 marks; 2025 · Paper-I · Q4(c) · 8 marks; 2026 · Paper-I · Q3(a) · 6 marks

Evidence limit: this page preserves the question records found in your Notion notes. Disputed labels and questions not checked against original PDFs remain flagged in the Question Bank; this is not a claim of exhaustive original-paper verification.