Scope of variables in C++ - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a variable inside the main function.

C++

int main() {
    int [1] = 10;
    return 0;
}

Drag options to blanks, or click blank then click option'

Anumber

Bcout

Cmain

Dreturn

Attempts:

3 left

💡 Hint

Common Mistakes

Using a keyword as a variable name.

Using function names as variable names.

The variable name inside main should be a valid identifier like number.

2fill in blank

medium

Complete the code to print the value of the local variable inside the function.

C++

#include <iostream>
using namespace std;

void show() {
    int x = 5;
    cout << [1] << endl;
}

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

Drag options to blanks, or click blank then click option'

Ax

By

Cz

Dcout

Attempts:

3 left

💡 Hint

Common Mistakes

Trying to print a variable not declared in the function.

Using variable names not declared anywhere.

The variable x is declared inside show() and can be printed there.

3fill in blank

hard

Fix the error by completing the code to access the global variable inside the function.

C++

#include <iostream>
using namespace std;

int [1] = 100;

void printGlobal() {
    cout << [1] << endl;
}

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

Drag options to blanks, or click blank then click option'

Anum

Bvalue

Cx

DglobalVar

Attempts:

3 left

💡 Hint

Common Mistakes

Using different names for the global variable declaration and usage.

Trying to access a variable not declared globally.

The global variable name must be the same in declaration and usage. Here, globalVar is used.

4fill in blank

hard

Fill both blanks to declare a variable inside a block and print it.

C++

#include <iostream>
using namespace std;

int main() {
    {
        int [1] = 20;
        cout << [2] << endl;
    }
    return 0;
}

Drag options to blanks, or click blank then click option'

AblockVar

CmainVar

Dvalue

Attempts:

3 left

💡 Hint

Common Mistakes

Using different names for declaration and printing.

Trying to access the variable outside the block.

The variable declared inside the block is blockVar, which is also printed inside the same block.

5fill in blank

hard

Fill all three blanks to declare a global variable, shadow it inside main, and print both values.

C++

#include <iostream>
using namespace std;

int [1] = 50;

int main() {
    int [2] = 100;
    cout << ::[3] << endl; // global
    cout << [2] << endl; // local
    return 0;
}

Drag options to blanks, or click blank then click option'

Acount

Dvalue

Attempts:

3 left

💡 Hint

Common Mistakes

Using different names for global and local variables.

Not using :: to access the global variable inside main.

The global variable count is declared outside main. Inside main, a local variable count shadows the global one. Using ::count accesses the global variable.

Practice

1. What is the scope of a variable declared inside a function in C++?

easy

A. It can only be used inside that function.

B. It can be used anywhere in the program.

C. It can be used only in the main function.

D. It can be used in all functions after declaration.

Solution

1. Step 1: Understand local variable scope

   A variable declared inside a function is local to that function only.

2. Step 2: Check usage outside the function

   Such variables cannot be accessed outside their function block.

3. Final Answer:

   It can only be used inside that function. -> Option A

4. Quick Check:

   Local variable scope = inside function only [OK]

Hint: Local variables live only inside their function block [OK]

Common Mistakes:

• Thinking local variables are accessible globally
• Confusing local and global variable scopes

2. Which of the following is the correct way to declare a global variable in C++?

easy

A. int x; inside main() function

B. int x; outside all functions

C. int x; inside a block { }

D. int x; inside a for loop

Solution

1. Step 1: Identify global variable declaration

   Global variables must be declared outside all functions and blocks.

2. Step 2: Check each option

   Only int x; outside all functions declares the variable outside all functions, making it global.

3. Final Answer:

   int x; outside all functions -> Option B

4. Quick Check:

   Global variable declared outside functions [OK]

Hint: Global variables declared outside all functions [OK]

Common Mistakes:

• Declaring global variables inside functions
• Declaring variables inside loops thinking they are global

3. What will be the output of the following C++ code?

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

medium

A. 10\n5

B. 10\n10

C. 5\n5

D. 5\n10

Solution

1. Step 1: Analyze variable shadowing in func()

   Inside func(), local x = 10 shadows global x = 5, so prints 10.

2. Step 2: Analyze main() output

   After func(), main() prints global x = 5.

3. Final Answer:

   10\n5 -> Option A

4. Quick Check:

   Local shadows global, prints 10 then 5 [OK]

Hint: Local variable shadows global inside function [OK]

Common Mistakes:

• Assuming global x changes inside func()
• Confusing which x is printed in main()

4. Identify the error in the following C++ code:

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

medium

A. Error: x declared twice

B. No error, prints 5

C. Error: Missing semicolon

D. Error: x used before declaration

Solution

1. Step 1: Check variable usage before declaration

   The code tries to print x before it is declared.

2. Step 2: Understand C++ variable initialization rules

   Variables must be declared before use; using x before declaration causes error.

3. Final Answer:

   Error: x used before declaration -> Option D

4. Quick Check:

   Use after declaration required [OK]

Hint: Declare variables before using them [OK]

Common Mistakes:

• Assuming declaration order does not matter
• Ignoring compiler errors about undeclared variables

5. Consider the following code snippet:

int x = 1;
void update() {
    x = 5;
    int x = 10;
    std::cout << x << std::endl;
}
int main() {
    update();
    std::cout << x << std::endl;
    return 0;
}

What will be the output and why?

hard

A. 10\n5 because local x shadows global, but global x updated before local declaration

B. 10\n1 because global x not updated due to local shadowing

C. Compilation error due to variable redeclaration

D. 5\n10 because global x updated first, then local x printed

Solution

1. Step 1: Analyze variable declaration order

   The line x = 5; appears before the local variable x is declared in the same function.

2. Step 2: Understand C++ rules on variable usage

   Using a variable before its local declaration in the same scope is illegal and causes a compilation error.

3. Step 3: Reason about the error

   Because local x is declared after x = 5;, the assignment tries to assign to local x before its declaration, which is invalid.

4. Final Answer:

   Compilation error due to variable redeclaration -> Option C

5. Quick Check:

   Using variable before declaration in same scope causes error [OK]

Hint: Cannot use variable before local declaration in same scope [OK]

Common Mistakes:

• Thinking there's a compilation error on x=5
• Believing local shadowing prevents global update before declaration