Stack overflow detection in Embedded C - Time & Space Complexity

Time Complexity: Stack overflow detection

O(1)

Understanding Time Complexity

We want to understand how the time needed to detect a stack overflow changes as the program runs.

How does the detection process scale when the stack grows larger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

#define STACK_SIZE 100
int stack[STACK_SIZE];
int top = -1;

int push(int value) {
  if (top >= STACK_SIZE - 1) {
    return -1; // Stack overflow detected
  }
  stack[++top] = value;
  return 0;
}
    

This code tries to add a value to a stack and checks if the stack is full to detect overflow.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: A single comparison to check if the stack is full before pushing.
• How many times: Once per push operation, no loops or recursion involved.

How Execution Grows With Input

Each push checks the top index against the stack size once, regardless of how many items are in the stack.

Input Size (n)	Approx. Operations
10	10 comparisons
100	100 comparisons
1000	1000 comparisons

Pattern observation: The number of operations grows directly with the number of push attempts, but each check is simple and constant time.

Final Time Complexity

Time Complexity: O(1)

This means each stack overflow check takes the same small amount of time no matter how big the stack is.

Common Mistake

[X] Wrong: "Checking for stack overflow takes longer as the stack grows."

[OK] Correct: The check only compares the top index to a fixed size, so it always takes the same time.

Interview Connect

Understanding that simple checks like stack overflow detection run in constant time helps you explain efficient code and avoid unnecessary loops.

Self-Check

"What if the stack overflow check involved scanning the entire stack array? How would the time complexity change?"

Practice

1.

What is the main purpose of stack overflow detection in embedded C?

easy

A. To increase the stack size dynamically

B. To prevent writing data beyond the stack's fixed size

C. To optimize CPU speed

D. To clear the stack memory automatically

Solution

1. Step 1: Understand stack overflow concept

   Stack overflow occurs when data is pushed beyond the stack's allocated size.

2. Step 2: Identify purpose of detection

   Detection helps prevent writing beyond stack limits, avoiding crashes and corruption.

3. Final Answer:

   To prevent writing data beyond the stack's fixed size -> Option B

4. Quick Check:

   Stack overflow detection = prevent overflow [OK]

Hint: Stack overflow detection stops writes beyond stack size [OK]

Common Mistakes:

• Thinking stack size can grow automatically
• Confusing overflow detection with CPU optimization
• Assuming stack clears itself on overflow

2.

Which of the following code snippets correctly checks for stack overflow before pushing a value?

int stack[10];
int sp = 0; // stack pointer
int value = 5;

// Check before push
if (____) {
    stack[sp++] = value;
}

easy

A. sp < 10

B. sp > 10

C. sp == 10

D. sp >= 10

Solution

1. Step 1: Understand stack pointer and size

   Stack size is 10, valid indices are 0 to 9, so sp must be less than 10 to push.

2. Step 2: Choose correct condition

   Condition 'sp < 10' ensures we don't write beyond array bounds.

3. Final Answer:

   sp < 10 -> Option A

4. Quick Check:

   Push only if sp less than stack size [OK]

Hint: Check sp < stack size before push [OK]

Common Mistakes:

• Using sp >= 10 allows overflow
• Using sp == 10 pushes out of bounds
• Using sp > 10 misses last valid index

3.

What will be the output of the following code snippet?

int stack[3];
int sp = 0;

void push(int val) {
    if (sp < 3) {
        stack[sp++] = val;
        printf("Pushed %d\n", val);
    } else {
        printf("Stack Overflow!\n");
    }
}

int main() {
    push(1);
    push(2);
    push(3);
    push(4);
    return 0;
}

medium

A. Pushed 1
Pushed 2
Pushed 3
Stack Overflow!

B. Pushed 1
Pushed 2
Pushed 3
Pushed 4

C. Stack Overflow!
Stack Overflow!
Stack Overflow!
Stack Overflow!

D. Pushed 1
Pushed 2
Stack Overflow!
Stack Overflow!

Solution

1. Step 1: Trace push calls and stack pointer

   First three pushes succeed because sp < 3; fourth push fails as sp == 3.

2. Step 2: Check output for each push

   Outputs "Pushed 1", "Pushed 2", "Pushed 3", then "Stack Overflow!" on last push.

3. Final Answer:

   Pushed 1 Pushed 2 Pushed 3 Stack Overflow! -> Option A

4. Quick Check:

   Pushes succeed until sp == 3, then overflow [OK]

Hint: Pushes stop when sp reaches stack size [OK]

Common Mistakes:

• Assuming push always succeeds
• Ignoring else branch output
• Confusing sp increment timing

4.

Identify the bug in the following stack push function that causes stack overflow detection to fail:

void push(int val) {
    if (sp <= 3) {
        stack[sp++] = val;
    } else {
        printf("Overflow detected\n");
    }
}

medium

A. sp is not incremented after push

B. Stack size is not checked at all

C. Condition should be sp < 3, not sp <= 3

D. printf statement is inside if block

Solution

1. Step 1: Analyze condition sp <= 3 with stack size 3

   sp can be 3, but valid indices are 0,1,2; sp=3 causes out-of-bounds write.

2. Step 2: Correct condition to prevent overflow

   Use sp < 3 to allow only valid indices 0 to 2.

3. Final Answer:

   Condition should be sp < 3, not sp <= 3 -> Option C

4. Quick Check:

   Use strict less than to avoid overflow [OK]

Hint: Use sp < stack size, not <=, to avoid overflow [OK]

Common Mistakes:

• Allowing sp == stack size causes overflow
• Not incrementing sp after push
• Misplacing overflow message

5.

You have a stack of size 5 and want to implement a safe push function that returns 0 on success and -1 on overflow. Which code snippet correctly implements this?

int stack[5];
int sp = 0;

int push(int val) {
    // Fill in condition
    if (__________) {
        stack[sp++] = val;
        return 0;
    } else {
        return -1;
    }
}

hard

A. sp > 5

B. sp <= 5

C. sp == 5

D. sp < 5

Solution

1. Step 1: Understand stack size and valid indices

   Stack size is 5, so valid indices are 0 to 4; sp must be less than 5 to push safely.

2. Step 2: Choose condition that prevents overflow

   Condition 'sp < 5' ensures no out-of-bounds write and returns 0 on success.

3. Final Answer:

   sp < 5 -> Option D

4. Quick Check:

   Push only if sp less than stack size [OK]

Hint: Return error if sp not less than stack size [OK]

Common Mistakes:

• Using sp <= 5 allows overflow
• Returning success on overflow
• Checking sp == 5 instead of less than