Array size and bounds in C - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does the size of an array represent in C?

The size of an array in C represents the total number of elements it can hold. For example, int arr[5]; means the array can hold 5 integers.

beginner

What happens if you access an array element outside its bounds in C?

Accessing an element outside the array bounds leads to undefined behavior. This means the program might crash, show wrong data, or behave unpredictably.

intermediate

How do you find the number of elements in an array in C?

You can find the number of elements by dividing the total size of the array by the size of one element: sizeof(arr) / sizeof(arr[0]).

beginner

Why is it important to stay within array bounds?

Staying within array bounds prevents bugs and crashes. It ensures your program accesses only valid memory, keeping data safe and the program stable.

beginner

What is the index range for an array declared as int arr[10];?

The valid index range is from 0 to 9. Arrays in C start at index 0, so the last element is at index size-1.

If you declare char letters[4];, what is the highest valid index you can use?

A 3

B 4

C 5

D 0

Array indices start at 0, so for 4 elements, valid indices are 0, 1, 2, and 3.

What does sizeof(arr) / sizeof(arr[0]) calculate for an array arr?

A Memory address of the array

B Size of one element

C Total bytes used by the array

D Number of elements in the array

Dividing total size by element size gives the number of elements.

What is the result of accessing arr[10] if int arr[10]; is declared?

A Last element of the array

B Undefined behavior

C Value 0

D Compile-time error

Accessing outside the array bounds causes undefined behavior, not a compile error.

Which of these is a valid array declaration for 5 integers?

A int arr();

B int arr;

C int arr[5];

D int arr{};

Only int arr[5]; declares an array of 5 integers.

Why should you avoid accessing array elements beyond their declared size?

A To prevent program crashes and data corruption

B To make the program run faster

C To save memory

D To increase array size automatically

Accessing out of bounds can cause crashes or corrupt data.

Explain what array bounds mean and why they are important in C programming.

Think about how arrays store data and what happens if you try to use invalid positions.

Describe how to calculate the number of elements in a C array using the sizeof operator.

Remember sizeof gives size in bytes.

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