Scope of variables in C++ - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of variable shadowing in nested blocks

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

C++

#include <iostream>

int main() {
    int x = 5;
    {
        int x = 10;
        std::cout << x << std::endl;
    }
    std::cout << x << std::endl;
    return 0;
}

A 10\n5\n

B 5\n10\n

C 10\n10\n

D 5\n5\n

💡 Hint

Remember that inner blocks can have variables with the same name that hide outer variables.

The inner block declares a new variable x with value 10, which shadows the outer x. So inside the block, x is 10. Outside, it remains 5.

❓ Predict Output

intermediate

Output when modifying a global variable inside a function

What is the output of this C++ program?

C++

#include <iostream>

int count = 0;

void increment() {
    count += 1;
}

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

A

2

B

0

C

1

D

Compilation error

💡 Hint

Global variables can be accessed and modified inside functions unless shadowed.

The global variable count starts at 0. Each call to increment() adds 1. After two calls, count is 2.

🔧 Debug

advanced

Identify the error caused by variable scope

What error does this code produce when compiled?

C++

#include <iostream>

int main() {
    if (true) {
        int a = 10;
    }
    std::cout << a << std::endl;
    return 0;
}

A Output: 10

B Runtime error: segmentation fault

C Compilation error: 'a' was not declared in this scope

D Compilation error: redefinition of 'a'

💡 Hint

Variables declared inside a block are not visible outside that block.

The variable a is declared inside the if block and is not accessible outside. Trying to use it outside causes a compilation error.

❓ Predict Output

advanced

Output of static variable inside a function

What is the output of this program?

C++

#include <iostream>

void counter() {
    static int count = 0;
    count++;
    std::cout << count << std::endl;
}

int main() {
    counter();
    counter();
    counter();
    return 0;
}

A

Compilation error

B

0
1
2

C

1
1
1

D

1
2
3

💡 Hint

Static variables inside functions keep their value between calls.

The static variable count is initialized once to 0, then incremented each call. So outputs 1, 2, 3.

🧠 Conceptual

expert

Understanding variable scope and lifetime in nested functions

Consider the following C++ code using lambda functions. What is the output?

C++

#include <iostream>
#include <functional>

int main() {
    int x = 5;
    auto f = [x]() mutable {
        x += 10;
        std::cout << x << std::endl;
    };
    f();
    f();
    std::cout << x << std::endl;
    return 0;
}

A

15
15
5

B

15
25
5

C

15
25
25

D

Compilation error

💡 Hint

Mutable lambdas capture variables by value but allow modification of the copy inside the lambda.

The lambda captures x by value, so it works on a copy stored in the lambda object. The mutable keyword allows modifying that copy, which persists across calls, outputting 15 then 25. The original x outside remains 5.