Auto storage class - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of auto variable inside a function

What is the output of this C program?

C

#include <stdio.h>

void func() {
    auto int x = 5;
    x++;
    printf("%d ", x);
}

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

A 6 6

B 6 7

C 5 6

D 5 5

💡 Hint

Remember that 'auto' variables inside functions are reinitialized each call.

The variable 'x' is declared as auto inside func(), so it is created anew each time func() is called. It starts at 5, then increments to 6, and prints 6 each time.

❓ Predict Output

intermediate

Effect of auto keyword on global variables

What error or output does this C code produce?

C

auto int x = 10;

int main() {
    printf("%d", x);
    return 0;
}

A 0

B 10

C Compilation error: 'auto' not allowed for global variables

D Undefined behavior

💡 Hint

Check if 'auto' can be used outside functions.

The 'auto' storage class is only valid for local variables inside functions. Using it for global variables causes a compilation error.

🔧 Debug

advanced

Identify the error with auto variable redeclaration

What error does this code produce?

C

int main() {
    auto int a = 5;
    auto int a = 10;
    printf("%d", a);
    return 0;
}

A Compilation error: redeclaration of 'a'

B 5

C 10

D Undefined behavior

💡 Hint

Can you declare the same variable twice in the same scope?

Declaring 'a' twice in the same scope causes a compilation error due to redeclaration.

🧠 Conceptual

advanced

Behavior of auto variables in loops

What is the output of this program?

C

#include <stdio.h>

int main() {
    for (int i = 0; i < 3; i++) {
        auto int x = 0;
        x += i;
        printf("%d ", x);
    }
    return 0;
}

A 0 1 3

B 0 1 2

C 0 0 0

D 1 2 3

💡 Hint

Each loop iteration creates a new 'x' starting at 0.

The variable 'x' is auto inside the loop, so it resets to 0 each iteration, then adds i, printing 0, 1, 2.

❓ Predict Output

expert

Auto variable and pointer behavior

What is the output of this C program?

C

#include <stdio.h>

int* func() {
    auto int x = 10;
    return &x;
}

int main() {
    int* p = func();
    printf("%d", *p);
    return 0;
}

A 10

B 0

C Compilation error

D Garbage value (undefined behavior)

💡 Hint

Think about the lifetime of 'x' after func() returns.

The function returns the address of an auto variable 'x' which is destroyed after the function ends. Accessing it causes undefined behavior and likely garbage output.

Practice

1. What does the auto storage class specify in C?

easy

A. The variable is stored in global memory and accessible everywhere.

B. The variable is stored in static memory and retains value between calls.

C. The variable is local to the function or block and has automatic lifetime.

D. The variable is stored in dynamic memory allocated at runtime.

Solution

1. Step 1: Understand the meaning of auto in C

   The auto keyword makes a variable local to the block or function where it is declared.

2. Step 2: Recognize the lifetime of auto variables

   Such variables exist only while the function or block runs, then they are destroyed.

3. Final Answer:

   The variable is local to the function or block and has automatic lifetime. -> Option C

4. Quick Check:

   auto = local, automatic lifetime [OK]

Hint: Remember: auto means local and temporary [OK]

Common Mistakes:

• Confusing auto with static storage
• Thinking auto variables are global
• Assuming auto variables persist after function ends

2. Which of the following is the correct syntax to declare an auto variable in C?

easy

A. int count auto = 5;

B. int auto count = 5;

C. auto = int count 5;

D. auto int count = 5;

Solution

1. Step 1: Recall the correct order of storage class and type

   In C, the storage class specifier auto comes before the type name.

2. Step 2: Check each option's syntax

   auto int count = 5; uses auto int count = 5; which is correct syntax. Others have incorrect order or invalid syntax.

3. Final Answer:

   auto int count = 5; -> Option D

4. Quick Check:

   Storage class before type = correct syntax [OK]

Hint: Storage class comes before type in declaration [OK]

Common Mistakes:

• Placing auto after the type
• Using assignment without type
• Incorrect order of keywords

3. What will be the output of this C code?

void func() {
    auto int x = 10;
    x++;
    printf("%d ", x);
}

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

medium

A. 11 12

B. 11 11

C. 10 10

D. 12 12

Solution

1. Step 1: Understand the behavior of auto variables in function

   The variable x is declared as auto int, so it is local and reinitialized each time func() runs.

2. Step 2: Trace the function calls

   Each call sets x = 10, then increments to 11, and prints 11. So output is "11 11 "

3. Final Answer:

   11 11 -> Option B

4. Quick Check:

   auto variable resets each call = 11 11 [OK]

Hint: auto resets each call, so output repeats [OK]

Common Mistakes:

• Assuming x keeps incrementing across calls
• Confusing auto with static variables
• Expecting output to be 11 12

4. Identify the error in this code snippet:

void example() {
    auto int count = 5;
    static int count = 10;
    printf("%d", count);
}

medium

A. Cannot declare two variables with the same name in the same scope.

B. auto and static cannot be used together in the same declaration.

C. auto variables cannot be initialized.

D. printf syntax is incorrect.

Solution

1. Step 1: Check variable declarations in the same scope

   Both count variables are declared in the same function scope, which is not allowed.

2. Step 2: Understand storage class rules

   Declaring two variables with the same name in the same scope causes a redeclaration error regardless of storage class.

3. Final Answer:

   Cannot declare two variables with the same name in the same scope. -> Option A

4. Quick Check:

   Same name in one scope = error [OK]

Hint: No duplicate variable names in one scope [OK]

Common Mistakes:

• Thinking auto and static conflict in one declaration
• Believing auto variables can't be initialized
• Ignoring variable redeclaration rules

5. Consider this code:

void test() {
    auto int a = 1;
    {
        auto int a = 2;
        printf("%d ", a);
    }
    printf("%d", a);
}

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

What is the output and why?

hard

A. 2 1 - Inner block has its own auto variable hiding outer one.

B. 1 1 - Outer variable is used both times.

C. 2 2 - Inner variable changes outer variable.

D. Compilation error due to duplicate variable names.

Solution

1. Step 1: Analyze variable scopes

   The inner block declares a new auto int a = 2; which hides the outer a inside the block.

2. Step 2: Trace the print statements

   Inside the block, a is 2, so prints "2 ". Outside, the outer a is still 1, so prints "1".

3. Final Answer:

   2 1 - Inner block has its own auto variable hiding outer one. -> Option A

4. Quick Check:

   Inner auto hides outer auto = 2 1 [OK]

Hint: Inner auto shadows outer variable in nested blocks [OK]

Common Mistakes:

• Assuming inner variable changes outer variable
• Expecting same variable used both times
• Thinking duplicate names cause error in nested blocks