Array size and bounds in C - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of accessing array within bounds

What is the output of this C program?

C

#include <stdio.h>

int main() {
    int arr[5] = {10, 20, 30, 40, 50};
    printf("%d\n", arr[3]);
    return 0;
}

A 40

B Runtime error

C 30

D 50

💡 Hint

Remember array indexing starts at 0.

The array has elements indexed 0 to 4. arr[3] is the fourth element, which is 40.

❓ Predict Output

intermediate

Output when accessing array out of bounds

What happens when this C code runs?

C

#include <stdio.h>

int main() {
    int arr[3] = {1, 2, 3};
    printf("%d\n", arr[5]);
    return 0;
}

A Undefined behavior (may print garbage or crash)

B Compile-time error

C 0

D 3

💡 Hint

Accessing outside array size is not safe in C.

Accessing arr[5] is beyond the declared size of 3, causing undefined behavior. It may print garbage or crash.

🧠 Conceptual

advanced

Size of array in bytes

Given this code, what is the output of the program?

C

#include <stdio.h>

int main() {
    int arr[10];
    printf("%zu\n", sizeof(arr));
    return 0;
}

A 10

B 40

C 4

D Depends on compiler, but usually 10

💡 Hint

sizeof returns size in bytes. int is usually 4 bytes.

The array has 10 integers. Each int is 4 bytes on most systems. So total size is 10 * 4 = 40 bytes.

📝 Syntax

advanced

Which array declaration is invalid?

Which of these array declarations will cause a compilation error in C?

A int arr[5];

B int arr[] = {1, 2, 3};

C int arr[1];

D int arr[3] = {1, 2, 3, 4};

💡 Hint

Array size must match initializer elements or be omitted.

Declaring int arr[3] with 4 initializers is invalid and causes a compile error.

🔧 Debug

expert

Why does this array access cause a crash?

Consider this code snippet. Why does it crash when run?

C

#include <stdio.h>

int main() {
    int arr[2] = {5, 10};
    for (int i = 0; i <= 2; i++) {
        printf("%d\n", arr[i]);
    }
    return 0;
}

A Because the loop condition should be i < 3 instead of i <= 2

B Because arr is not initialized properly

C Because the loop accesses arr[2], which is out of bounds causing undefined behavior

D Because printf format specifier is incorrect

💡 Hint

Check the loop condition and array size carefully.

The array has size 2, valid indices are 0 and 1. Accessing arr[2] is out of bounds and causes a crash.

Practice

1. What is the valid range of indexes for an array declared as int arr[10]; in C?

easy

A. 0 to 9

B. 1 to 10

C. 0 to 10

D. 1 to 9

Solution

1. Step 1: Understand array size declaration

   The array arr is declared with size 10, meaning it has 10 elements.

2. Step 2: Determine valid index range

   In C, array indexing starts at 0, so valid indexes are from 0 up to size-1, which is 9.

3. Final Answer:

   0 to 9 -> Option A

4. Quick Check:

   Array indexes = 0 to size-1 = 0 to 9 [OK]

Hint: Array indexes start at 0 and end at size minus one [OK]

Common Mistakes:

• Assuming array indexes start at 1
• Using size as the last index
• Confusing size with maximum index

2. Which of the following is the correct way to declare an array of 5 integers in C?

easy

A. int arr(5);

B. int arr[5];

C. int arr{5};

D. int arr = 5;

Solution

1. Step 1: Recall C array declaration syntax

   In C, arrays are declared using square brackets with the size inside, like int arr[5];.

2. Step 2: Check each option

   int arr[5]; uses correct syntax. Options A and C use parentheses or braces which are invalid for arrays. int arr = 5; assigns an int, not an array.

3. Final Answer:

   int arr[5]; -> Option B

4. Quick Check:

   Array declaration uses square brackets [OK]

Hint: Use square brackets [] to declare array size in C [OK]

Common Mistakes:

• Using parentheses instead of brackets
• Using braces instead of brackets
• Assigning a single value instead of declaring array

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

int arr[3] = {10, 20, 30};
printf("%d", arr[3]);

medium

A. Undefined behavior (may print garbage or crash)

B. 0

C. Compilation error

D. 30

Solution

1. Step 1: Analyze array size and indexing

   The array arr has size 3, valid indexes are 0, 1, 2. Accessing arr[3] is out of bounds.

2. Step 2: Understand consequences of out-of-bounds access

   Accessing outside array bounds in C causes undefined behavior. It may print garbage value, cause crash, or unpredictable output.

3. Final Answer:

   Undefined behavior (may print garbage or crash) -> Option A

4. Quick Check:

   Out-of-bounds access = undefined behavior [OK]

Hint: Accessing index >= size causes undefined behavior [OK]

Common Mistakes:

• Assuming arr[3] prints last element
• Expecting zero instead of garbage
• Thinking compiler catches out-of-bounds

4. Identify the error in the following code snippet:

int arr[4];
for(int i = 0; i <= 4; i++) {
    arr[i] = i * 2;
}

medium

A. Initialization of i is incorrect

B. Array size should be 5 instead of 4

C. Loop condition should be i < 4, not i <= 4

D. Multiplication by 2 is invalid

Solution

1. Step 1: Check loop bounds against array size

   The array size is 4, so valid indexes are 0 to 3. The loop runs while i <= 4, which includes i=4, an invalid index.

2. Step 2: Correct the loop condition

   Changing the loop condition to i < 4 ensures indexes 0 to 3 are used, avoiding out-of-bounds access.

3. Final Answer:

   Loop condition should be i < 4, not i <= 4 -> Option C

4. Quick Check:

   Loop index must be less than array size [OK]

Hint: Use < not <= for loop index to avoid out-of-bounds [OK]

Common Mistakes:

• Using <= instead of < in loop condition
• Assuming array size includes index equal to size
• Ignoring off-by-one errors

5. Given the array declaration int nums[5] = {1, 2, 3, 4, 5};, which code snippet correctly sums all elements without causing out-of-bounds errors?

hard

A. for(int i = 0; i < 6; i++) sum += nums[i];

B. for(int i = 1; i < 5; i++) sum += nums[i];

C. for(int i = 0; i <= 5; i++) sum += nums[i];

D. for(int i = 0; i < 5; i++) sum += nums[i];

Solution

1. Step 1: Understand array size and valid indexes

   The array nums has size 5, so valid indexes are 0 to 4.

2. Step 2: Evaluate each loop condition

   for(int i = 0; i <= 5; i++) sum += nums[i]; uses i <= 5, which accesses nums[5] (out-of-bounds). for(int i = 1; i < 5; i++) sum += nums[i]; starts at 1, missing nums[0]. for(int i = 0; i < 5; i++) sum += nums[i]; uses i < 5, correctly iterating 0 to 4. for(int i = 0; i < 6; i++) sum += nums[i]; uses i < 6, accessing nums[5] (out-of-bounds).

3. Final Answer:

   for(int i = 0; i < 5; i++) sum += nums[i]; -> Option D

4. Quick Check:

   Loop from 0 to size-1 to avoid errors [OK]

Hint: Loop from 0 to less than array size for safe access [OK]

Common Mistakes:

• Using <= size causing out-of-bounds
• Starting loop at 1 missing first element
• Looping beyond array size