Topic Revision Tests — T5 to T8

12 questions each · ~20 min each · span-units consolidation · attempt ALL before any key · ← T1–T4 · Hub · Cumulative tests →

T5 — Strings & text processing (Units 03, 11)

Q1. [Easy] Which correctly tests whether string s is empty?

• a) s == ""
• b) s.empty()
• c) s.size() == 0
• d) all three work

Q2. [Easy] string s = "Programming"; — s.substr(0, 7) returns…

• a) Program
• b) ming
• c) Programming
• d) Pro

Q3. [Easy] Concatenation of first = "Ada" and last = "Lovelace" with a space is…

• a) first + " " + last
• b) first ++ " " ++ last
• c) first.append(last)
• d) first.space(last)

Q4. [Medium] What does this print?

string s = "abc";
for (int i = s.size() - 1; i >= 0; i--) cout << s[i];

• a) abc
• b) cba
• c) c
• d) nothing

Q5. [Medium] What does this print?

string s = "Hello World";
int spaces = 0;
for (size_t i = 0; i < s.size(); i++)
    if (s[i] == ' ') spaces++;
cout << spaces << " " << s.size();

• a) 1 11
• b) 1 10
• c) 0 11
• d) 2 11

Q6. [Medium] After cin >> n; with input 42, what does getline(cin, line) (immediately after) store in line?

• a) the user’s next full line
• b) an empty string — the buffered newline satisfied it
• c) “42”
• d) it crashes

Q7. [Medium] What does this print?

string s = "2026";
int n = 0;
for (size_t i = 0; i < s.size(); i++) n = n * 10 + (s[i] - '0');
cout << n + 1;

• a) 2027 — the digit-to-value conversion builds the number
• b) 20261
• c) 0
• d) an error

Q8. [Medium] Which snippet capitalizes the first character of word?

• a) word[0] = toupper(word[0]);
• b) toupper(word);
• c) word.first() = upper;
• d) word = upper(word);

Q9. [Hard] What does this print?

string a = "apple", b = "Apple";
if (a == b)      cout << "same";
else if (a < b)  cout << "lower-first";
else             cout << "upper-first";

• a) same
• b) lower-first — uppercase letters precede lowercase in ASCII, so “Apple” < “apple”… check carefully
• c) upper-first
• d) undefined

Q10. [Hard] What does this print?

string s = "mississippi";
int count = 0;
for (size_t i = 0; i + 2 < s.size(); i++)
    if (s.substr(i, 2) == "ss") count++;
cout << count;

• a) 2
• b) 3 — overlapping occurrences counted: positions 2, 5, 8
• c) 1
• d) 4

Q11. [Hard] What does this print?

string line = "  data  ";
// strip leading and trailing spaces
size_t b = line.find_first_not_of(' ');
size_t e = line.find_last_not_of(' ');
cout << "[" << line.substr(b, e - b + 1) << "]";

• a) [data]
• b) [ data ]
• c) [data ]
• d) []

Q12. [Hard] A word counter must treat runs of spaces as one separator ("a b" is two words). The loop test that handles it is…

• a) count every space
• b) count a word-start: current char is not a space AND (i == 0 OR previous char was a space)
• c) split on every character
• d) count non-space characters

T6 — Files & persistence (Unit 12)

Q1. [Easy] Which stream type reads a file?

• a) ofstream
• b) ifstream
• c) fstream writing
• d) stringstream

Q2. [Easy] ofstream out("data.txt"); on an existing file…

• a) appends to it
• b) truncates it — old contents are gone
• c) fails if the file exists
• d) opens it read-only

Q3. [Easy] Which line correctly checks an open?

• a) if (in.open()) ...
• b) if (!in) { /* handle failure */ }
• c) if (in == true) ...
• d) check(in);

Q4. [Medium] nums.txt holds 10 20 30. What does this print?

ifstream in("nums.txt");
int x, sum = 0, count = 0;
while (in >> x) { sum += x; count++; }
cout << count << " " << sum;

• a) 3 60
• b) 1 60
• c) 0 0
• d) 3 102030

Q5. [Medium] What does this loop read?

string line;
while (getline(in, line)) { ... }

• a) one word per pass
• b) one full line per pass — including interior spaces, newline consumed
• c) only lines with numbers
• d) nothing; getline cannot loop

Q6. [Medium] A log must add lines across many program runs. The open is…

• a) ofstream log("log.txt");
• b) ofstream log("log.txt", ios::app);
• c) ifstream log("log.txt");
• d) fstream log("log.txt", ios::in);

Q7. [Medium] What does this print, given the file exists with 4 lines?

ifstream in("words.txt");
string w;
int count = 0;
while (in >> w) count++;
cout << count;

• a) 4 — the file has 4 lines, one word each
• b) 0
• c) 1
• d) unknown without the word count per line

Q8. [Medium] Writing out << name << "," << marks << "\n"; produces…

• a) a binary record
• b) one CSV-style line per student
• c) a compile error
• d) output only after program exit

Q9. [Hard] stock.txt holds Pen 3\nPad 990\n. What does this print?

ifstream in("stock.txt");
string name; int qty;
while (in >> name >> qty)
    if (qty < 10) cout << name << " ";

• a) Pen — the low-stock filter over streamed records
• b) Pad
• c) Pen Pad
• d) nothing

Q10. [Hard] A program reads a file, averages the numbers, and must also append the average as a final line. The safe structure is…

• a) read to EOF in the same stream, then immediately out << avg; on that stream after reopening for append — two opens, two purposes
• b) read and append interleaved on one ofstream
• c) append first, read later
• d) keep the ifstream and cast it to ofstream

Q11. [Hard] marks.txt holds Ayesha,88\nBilal,72\n. What does this print?

ifstream in("marks.txt");
string line;
int total = 0, count = 0;
while (getline(in, line)) {
    size_t comma = line.find(',');
    if (comma == string::npos) continue;
    total += stoi(line.substr(comma + 1));
    count++;
}
cout << total << "/" << count;

• a) 160/2
• b) 88/1
• c) 160/1
• d) nothing — stoi cannot read from files

Q12. [Hard] The deepest reason if (!in) after opening matters: a failed open…

• a) throws immediately, so the check is redundant
• b) leaves the stream in a fail state — every subsequent read is a no-op, and averages computed over zero records would print nan or lie
• c) terminates the program automatically
• d) only matters on Windows

T7 — Memory & pointers (Unit 13)

Q1. [Easy] Which stores the address of x?

• a) int p = x;
• b) int* p = &x;
• c) int& p = x;
• d) int p = *x;

Q2. [Easy] cout << *p; where p points at n prints…

• a) the address of n
• b) the value of n
• c) the address of p
• d) nothing

Q3. [Easy] nullptr is…

• a) a dereference error
• b) the pointer value meaning “points at nothing”
• c) a type of int
• d) the address of main

Q4. [Medium] What does this print?

int a = 3, b = 8;
int* p = &a;
p = &b;
cout << *p;

• a) 3
• b) 8 — the pointer was reseated to b
• c) 11
• d) the address of b

Q5. [Medium] What does this print?

int arr[3] = {10, 20, 30};
int* p = arr;
cout << *p << *(p + 1);

• a) 1020
• b) 10 20 without space — exactly 1020
• c) 102030
• d) addresses

Q6. [Medium] What does this print?

void viaPointer(int* p) { *p = *p + 1; }
int main() {
    int v = 41;
    viaPointer(&v);
    cout << v;
}

• a) 41
• b) 42 — the function wrote through the address
• c) 0
• d) the address of v

Q7. [Medium] What does this print?

int* data = new int[3];
data[0] = 5; data[1] = 6; data[2] = 7;
cout << data[1];
delete[] data;

• a) 5
• b) 6
• c) 7
• d) an error — delete before printing

Q8. [Medium] Which is a memory leak?

• a) int* p = new int; delete p;
• b) int* p = new int; with no delete before p goes out of scope
• c) int x = 5;
• d) delete nullptr; (safe, does nothing)

Q9. [Hard] What does this print?

int a[4] = {2, 4, 6, 8};
int* end = a + 4;
int s = 0;
for (int* p = a; p < end; p++) s += *p;
cout << s;

• a) 20 — the pointer-walk sum
• b) 8
• c) 14
• d) an error — a + 4 is invalid

Q10. [Hard] What does this print?

int x = 1;
int* p = &x;
int** pp = &p;
**pp = 99;
cout << x;

• a) 1
• b) 99 — two dereferences reach x
• c) the address of x
• d) undefined

Q11. [Hard] After delete[] data;, the variable data…

• a) becomes nullptr automatically
• b) still holds the old address — now dangling; reuse is undefined behavior
• c) points at a zeroed array
• d) is a compile error

Q12. [Hard] int& r = x; versus int* p = &x; — the deepest difference…

• a) none; both are addresses
• b) a reference is a permanent alias bound once (cannot be reseated, never null, used without *); a pointer is a re-pointable object holding an address
• c) references are slower
• d) pointers cannot point to ints

T8 — OOP & design (Units 14–15)

Q1. [Easy] class differs from struct only in…

• a) available features
• b) default member access: private for class, public for struct
• c) memory layout
• d) compilation speed

Q2. [Easy] A constructor…

• a) is named init
• b) shares the class name and runs at object creation
• c) must return void
• d) is optional for every class — always optional, never required… unless initialization is needed

Q3. [Easy] Encapsulation chiefly means…

• a) putting everything in one file
• b) private state plus a public interface that enforces the class’s rules
• c) fast execution
• d) using structs only

Q4. [Medium] What does this print?

class Meter {
    int v;
public:
    Meter() : v(0) {}
    void up()   { v++; }
    void down() { if (v > 0) v--; }   // the invariant: never negative
    int  get() const { return v; }
};
int main() {
    Meter m;
    m.down(); m.up(); m.up();
    cout << m.get();
}

• a) 1 — down() was refused at 0 by the guard
• b) 2
• c) −1
• d) 0

Q5. [Medium] What does this print?

class P {
public:
    P()  { cout << "C"; }
    ~P() { cout << "D"; }
};
void make() { P p; }
int main() {
    make();
    cout << "!";
}

• a) CD!
• b) CD!D — no: only one object exists
• c) CD!
• d) C!D

Q6. [Medium] What does this print?

class Box {
    int side;
public:
    Box(int s) : side(s) {}
    int area() const { return side * side; }
};
void show(const Box& b) { cout << b.area(); }
int main() {
    Box b(4);
    show(b);
}

• a) 16 — const& accepted; area() is const, callable on it
• b) compile error: area not const
• c) 4
• d) 0

Q7. [Medium] What does this print?

class Chain {
    int v = 1;
public:
    Chain& twice() { v *= 2; return *this; }
    Chain& inc()   { v += 1; return *this; }
    int get() const { return v; }
};
int main() {
    Chain c;
    cout << c.twice().twice().inc().get();
}

• a) 5 — 1→2→4→5 via the returned *this
• b) 4
• c) 8
• d) 3

Q8. [Medium] Composition — a Car containing an Engine member — is preferred over inheritance when the relationship is…

• a) is-a
• b) has-a
• c) both
• d) neither

Q9. [Hard] What does this print?

class Stack {
    vector<int> data;
public:
    void push(int x) { data.push_back(x); }
    int pop() {
        int top = data.back();
        data.pop_back();
        return top;
    }
};
int main() {
    Stack s;
    s.push(1); s.push(2); s.push(3);
    cout << s.pop() << s.pop();
}

• a) 32 — LIFO: the last push comes off first
• b) 12
• c) 33
• d) 23

Q10. [Hard] What does this print?

class Item {
public:
    string name;
    int qty;
    Item(string n, int q) : name(n), qty(q) {}
};
ostream& operator<<(ostream& os, const Item& it) {
    return os << it.name << ":" << it.qty;
}
int main() {
    Item it("bolt", 40);
    cout << it;
}

• a) bolt:40
• b) Item
• c) an address
• d) a compile error

Q11. [Hard] A setter must reject a marks value outside 0–100. The design that keeps the invariant safest is…

• a) public data member; callers are trusted
• b) private data + a setter that refuses invalid values (bool return or throw) + a constructor using the same path
• c) validation only in the printing function
• d) a global validity flag

Q12. [Hard] The destructor’s defining property is…

• a) it must be called manually
• b) it runs automatically when the object’s lifetime ends — the hook RAII uses for cleanup
• c) it returns the object’s memory cost
• d) it can be overloaded per argument

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