Type modifiers in C++ - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of const pointer vs pointer to const

What is the output of the following C++ code?

C++

#include <iostream>

int main() {
    int x = 5;
    int y = 10;
    const int* ptr1 = &x;
    int* const ptr2 = &y;

    // *ptr1 = 6; // Line A (commented out)
    ptr1 = &y;    // Line B

    *ptr2 = 15;   // Line C
    // ptr2 = &x;  // Line D (commented out)

    std::cout << "*ptr1 = " << *ptr1 << ", *ptr2 = " << *ptr2 << std::endl;
    return 0;
}

A *ptr1 = 5, *ptr2 = 15

B *ptr1 = 5, *ptr2 = 10

C *ptr1 = 10, *ptr2 = 10

D *ptr1 = 10, *ptr2 = 15

💡 Hint

Remember that const int* ptr means the value pointed to is constant, but the pointer itself can change. int* const ptr means the pointer is constant, but the value pointed to can change.

const int* ptr1 means you cannot change the value through ptr1, but you can change where ptr1 points. So ptr1 = &y; is valid.

int* const ptr2 means the pointer itself cannot change, but the value it points to can be changed. So *ptr2 = 15; is valid, but ptr2 = &x; is not.

Therefore, *ptr1 points to y which is 10, and *ptr2 is changed to 15.

❓ Predict Output

intermediate

Effect of volatile modifier on variable

What will be the output of this C++ program?

C++

#include <iostream>

int main() {
    volatile int a = 10;
    int b = 10;

    a = a + 5;
    b = b + 5;

    std::cout << "a = " << a << ", b = " << b << std::endl;
    return 0;
}

A a = 10, b = 15

B a = 15, b = 15

C a = 15, b = 10

D Compilation error due to volatile

💡 Hint

The volatile keyword tells the compiler not to optimize access to the variable, but it does not prevent normal operations.

The volatile modifier means the variable can be changed unexpectedly (e.g., by hardware), so the compiler must always read it fresh. However, normal arithmetic and assignment work as usual. So both a and b become 15.

❓ Predict Output

advanced

Output of signed and unsigned arithmetic

What is the output of this C++ code snippet?

C++

#include <iostream>

int main() {
    unsigned int u = 5;
    int s = -10;
    auto result = u + s;
    std::cout << result << std::endl;
    return 0;
}

A Compilation error due to mixed types

B -5

C 4294967291

D 5

💡 Hint

When mixing signed and unsigned integers, the signed integer is converted to unsigned before the operation.

The signed integer -10 is converted to unsigned (assuming 32-bit unsigned int) which becomes 4294967286. Adding 5 results in 4294967291.

Note: The exact output depends on the system's unsigned int size, but typically 32-bit.

❓ Predict Output

advanced

Effect of mutable keyword in const object

What will be the output of this C++ program?

C++

#include <iostream>

class Counter {
public:
    mutable int count = 0;
    void increment() const {
        count++;
    }
};

int main() {
    const Counter c;
    c.increment();
    std::cout << c.count << std::endl;
    return 0;
}

A 1

B 0

C Compilation error due to modifying const object

D Undefined behavior

💡 Hint

The mutable keyword allows modification of a member even in a const object.

The member count is declared mutable, so it can be changed inside a const method and even when the object is const. The increment() method is const but modifies count. So after calling increment(), count becomes 1.

🧠 Conceptual

expert

Which option causes a compilation error due to type modifiers?

Which of the following code snippets will cause a compilation error because of incorrect use of type modifiers?

A const int* p = nullptr; *p = 10;

B volatile int v = 10; v = 20;

C int* const p = new int(5);

D const int x = 5; int* p = (int*)&x; *p = 10;

💡 Hint

Think about what const int* means and whether you can modify the value it points to.

Option A tries to modify the value pointed to by a const int*, which is not allowed and causes a compilation error.

Option A is valid; volatile does not prevent assignment.

Option A is valid; a constant pointer (int* const) can be initialized at declaration.

Option A uses a cast to remove constness and modifies a const variable, which is undefined behavior but compiles.

Practice

1. Which of the following type modifiers in C++ makes a variable value unchangeable after initialization?

easy

A. long

B. unsigned

C. volatile

D. const

Solution

1. Step 1: Understand the role of const

   The const modifier makes a variable's value fixed after it is set.

2. Step 2: Compare with other modifiers

   unsigned changes number sign, volatile affects optimization, long changes size, but none prevent value change.

3. Final Answer:

   const -> Option D

4. Quick Check:

   const = unchangeable value [OK]

Hint: const means value cannot be changed after set [OK]

Common Mistakes:

• Confusing const with unsigned
• Thinking volatile prevents changes
• Assuming long affects constness

2. Which of the following declarations correctly uses the unsigned modifier for an integer variable named count?

easy

A. unsigned count int;

B. count unsigned int;

C. unsigned int count;

D. int count unsigned;

Solution

1. Step 1: Recall correct syntax for unsigned int

   The correct order is unsigned int followed by the variable name.

2. Step 2: Check each option

   unsigned count int; invalid order; count unsigned int; places variable before type, invalid; unsigned int count; correct; int count unsigned; invalid.

3. Final Answer:

   unsigned int count; -> Option C

4. Quick Check:

   unsigned int before variable name [OK]

Hint: Put unsigned before int and then variable name [OK]

Common Mistakes:

• Placing unsigned after variable name
• Mixing order of type and modifier
• Using invalid syntax like unsigned count int

3. What is the output of the following C++ code?

unsigned int x = 10;
int y = -5;
if (x + y > 5)
    std::cout << "True";
else
    std::cout << "False";

medium

A. True

B. False

C. Compilation error

D. Undefined behavior

Solution

1. Step 1: Understand unsigned and signed addition

   Adding signed -5 to unsigned 10: -5 promotes to unsigned (large value, e.g., 4294967291u on 32-bit).

2. Step 2: Calculate the result and compare

   10u + 4294967291u = 4294967286u (wraps around mod 2^32). 4294967286u > 5u is true, so prints "True".

3. Final Answer:

   True -> Option A

4. Quick Check:

   unsigned + negative int promotes and wraps to large unsigned > 5 true [OK]

Hint: Mixing unsigned + signed negative promotes and wraps around [OK]

Common Mistakes:

• Assuming -5 stays negative (ignores promotion)
• Expecting True (ignores wraparound)
• Thinking this causes compile error

4. Identify the error in this code snippet:

const int x;
x = 5;

medium

A. Missing initialization of const variable

B. Incorrect use of int type

C. Assignment operator is wrong

D. No error, code is correct

Solution

1. Step 1: Check const variable initialization rules

   A const variable must be initialized when declared.

2. Step 2: Analyze the code

   Here, const int x; is declared without initialization, then assigned later, which is invalid.

3. Final Answer:

   Missing initialization of const variable -> Option A

4. Quick Check:

   const must be initialized immediately [OK]

Hint: const variables need immediate initialization [OK]

Common Mistakes:

• Trying to assign const after declaration
• Ignoring initialization requirement
• Thinking assignment operator is wrong

5. You want to store a large positive number that can never be negative and might exceed 2 billion. Which type modifier combination is best for the variable score?

hard

A. const int score;

B. unsigned long score;

C. short unsigned score;

D. volatile int score;

Solution

1. Step 1: Understand the data requirements

   We need a large positive number, no negatives, possibly > 2 billion.

2. Step 2: Match type modifiers to needs

   unsigned ensures no negatives; long increases size to hold large values; const fixes value; short limits size; volatile affects optimization, not size.

3. Step 3: Choose best option

   unsigned long fits the need best.

4. Final Answer:

   unsigned long score; -> Option B

5. Quick Check:

   unsigned long = large positive number [OK]

Hint: Use unsigned long for large positive numbers [OK]

Common Mistakes:

• Choosing const which fixes value
• Using short which limits size
• Using volatile which doesn't affect size