One-dimensional arrays in C - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of array element modification

What is the output of this C program?

C

#include <stdio.h>

int main() {
    int arr[5] = {1, 2, 3, 4, 5};
    arr[2] = arr[0] + arr[4];
    printf("%d\n", arr[2]);
    return 0;
}

A 3

B 5

C 6

D 9

💡 Hint

Remember array indexing starts at 0. arr[2] is changed to arr[0] + arr[4].

arr[0] is 1 and arr[4] is 5, so arr[2] becomes 1 + 5 = 6.

❓ Predict Output

intermediate

Sum of elements in an array

What is the output of this C program?

C

#include <stdio.h>

int main() {
    int nums[] = {10, 20, 30, 40};
    int sum = 0;
    for(int i = 0; i < 4; i++) {
        sum += nums[i];
    }
    printf("%d\n", sum);
    return 0;
}

A 40

B 100

C 90

D 10

💡 Hint

Add all elements: 10 + 20 + 30 + 40.

The sum is 10 + 20 + 30 + 40 = 100.

🔧 Debug

advanced

Identify the error in array access

What error will this C program produce when run?

C

#include <stdio.h>

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

A Undefined behavior (out-of-bounds access)

B Compilation error: array index out of range

C Runtime error: segmentation fault

D Prints 0

💡 Hint

Array indices go from 0 to size-1. Accessing arr[3] is outside the array.

Accessing arr[3] is out of bounds and causes undefined behavior. The program may print garbage or crash, but no compile error occurs.

📝 Syntax

advanced

Which option causes a compilation error?

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

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

B int arr[] = {1, 2, 3, 4, 5};

C int arr[5] = {1, 2, 3, 4, 5};

D int arr[5];

💡 Hint

Check if the initializer list fits the declared array size.

Option A tries to initialize an array of size 3 with 4 elements, which is a compilation error.

🚀 Application

expert

Find the number of elements in the array

What is the value of 'count' after running this C program?

C

#include <stdio.h>

int main() {
    int data[] = {5, 10, 15, 20, 25, 30};
    int count = sizeof(data) / sizeof(data[0]);
    printf("%d\n", count);
    return 0;
}

A 4

B 5

C 24

D 6

💡 Hint

sizeof(data) gives total bytes, sizeof(data[0]) gives bytes per element.

The array has 6 elements. Dividing total size by element size gives 6.

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