One-dimensional arrays in C - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is a one-dimensional array in C?

A one-dimensional array in C is a collection of elements of the same type stored in contiguous memory locations. It is like a list where each element can be accessed by its index.

beginner

How do you declare a one-dimensional array of 5 integers in C?

You declare it like this: int arr[5]; This creates an array named arr that can hold 5 integers.

beginner

How do you access the third element of an array named arr?

You access it using the index 2: arr[2]. Remember, array indexes start at 0, so the first element is arr[0].

intermediate

What happens if you try to access an array element outside its declared size?

Accessing outside the array size causes undefined behavior. It may read or write memory it shouldn't, leading to bugs or crashes.

beginner

How can you initialize a one-dimensional array with values at the time of declaration?

You can write: int arr[3] = {10, 20, 30}; This sets the first element to 10, second to 20, and third to 30.

What is the index of the first element in a C array?

A 0

B 1

C -1

D Depends on the array size

In C, array indexing always starts at 0.

Which of the following declares an array of 10 floats?

A int arr[10];

B float arr[10];

C float arr;

D float arr(10);

The correct syntax for declaring an array of 10 floats is float arr[10];.

What will happen if you write to arr[5] when arr is declared as int arr[5];?

A It causes undefined behavior

B It writes to the last element safely

C It throws a compile-time error

D It automatically resizes the array

Accessing arr[5] is outside the bounds since valid indexes are 0 to 4, causing undefined behavior.

How do you initialize an array with all zeros in C?

A int arr[5] = NULL;

B int arr[5] = 0;

C int arr[5]; // automatically zeroed

D int arr[5] = {0};

Initializing with {0} sets the first element to zero and all others to zero by default.

Which statement correctly accesses the last element of int arr[7];?

A arr[-1]

B arr[7]

C arr[6]

D arr[0]

The last element index is size minus one, so for 7 elements, it is arr[6].

Explain what a one-dimensional array is and how you use it in C.

Think of it like a row of boxes where each box holds a value.

Describe how to declare, initialize, and access elements in a one-dimensional array in C.

Remember the array size and zero-based indexing.

Practice

1. What is the starting index of a one-dimensional array in C?

easy

A. 0

B. 1

C. -1

D. Depends on the array size

Solution

1. Step 1: Understand array indexing in C

   In C, arrays always start indexing from 0, meaning the first element is at index 0.

2. Step 2: Confirm the starting index

   This is a fixed rule in C and does not depend on array size or other factors.

3. Final Answer:

   0 -> Option A

4. Quick Check:

   Array indexing starts at 0 [OK]

Hint: Remember: C arrays always start at zero index [OK]

Common Mistakes:

• Thinking arrays start at 1 like in some other languages
• Assuming negative indexes are valid
• Believing index depends on array size

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

easy

A. int arr = 5;

B. int arr();

C. array int arr[5];

D. int arr[5];

Solution

1. Step 1: Recall C array declaration syntax

   In C, arrays are declared by specifying the type, name, and size in square brackets, like int arr[5];

2. Step 2: Check each option

   int arr[5]; matches the correct syntax. int arr(); uses parentheses which is for functions. int arr = 5; assigns a single int, not an array. array int arr[5]; uses invalid keyword 'array'.

3. Final Answer:

   int arr[5]; -> Option D

4. Quick Check:

   Array declaration uses square brackets [OK]

Hint: Use square brackets for array size in declaration [OK]

Common Mistakes:

• Using parentheses instead of square brackets
• Assigning a single value instead of declaring array
• Using invalid keywords like 'array'

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

int arr[3] = {10, 20, 30};
printf("%d", arr[1]);

medium

A. 10

B. 30

C. 20

D. Compilation error

Solution

1. Step 1: Identify the array elements and indexes

   The array arr has elements: arr[0]=10, arr[1]=20, arr[2]=30.

2. Step 2: Understand the printf statement

   printf prints the value at arr[1], which is 20.

3. Final Answer:

   20 -> Option C

4. Quick Check:

   arr[1] = 20 [OK]

Hint: Remember array index 1 is second element [OK]

Common Mistakes:

• Confusing index 1 with index 0
• Expecting output 10 or 30 incorrectly
• Thinking code causes compilation error

4. Find the error in this code snippet:

int arr[4];
arr[4] = 10;

medium

A. Array index out of bounds

B. Missing semicolon

C. Wrong data type

D. No error

Solution

1. Step 1: Check array size and indexing

   The array arr has size 4, so valid indexes are 0,1,2,3 only.

2. Step 2: Identify the invalid index usage

   arr[4] accesses the fifth element, which is outside the array bounds causing undefined behavior.

3. Final Answer:

   Array index out of bounds -> Option A

4. Quick Check:

   Index must be less than array size [OK]

Hint: Index must be less than array size [OK]

Common Mistakes:

• Using index equal to array size
• Ignoring zero-based indexing
• Assuming no error for out-of-bounds access

5. Given the array declaration int nums[5] = {1, 2, 3, 4, 5};, which code snippet correctly calculates the sum of all elements?

hard

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

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

C. int sum = 0;
for(int i = 1; i < 5; i++) sum += nums[i];

D. int sum = 0;
for(int i = 0; i < 4; i++) sum += nums[i];

Solution

1. Step 1: Understand array size and valid indexes

   The array nums has 5 elements with indexes 0 to 4.

2. Step 2: Analyze each loop condition

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

3. Final Answer:

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

4. Quick Check:

   Loop from 0 to less than size sums all elements [OK]

Hint: Loop from 0 to less than array size for all elements [OK]

Common Mistakes:

• Using <= size causing out-of-bounds access
• Starting loop at 1 missing first element
• Stopping loop early missing last element