Array Access and Update at Index in DSA C - Execution Trace

Concept Flow - Array Access and Update at Index

Start with array

↓

Access element at index i

↓

Read or update value

↓

Store updated value if any

↓

End operation

The flow shows starting with an array, accessing an element by index, optionally updating it, and then storing the new value.

Execution Sample

DSA C

int arr[5] = {10, 20, 30, 40, 50};
int x = arr[2];
arr[3] = 100;

This code reads the element at index 2 and updates the element at index 3.

Execution Table

Step	Operation	Index Accessed	Value Read	Value Updated	Array State
1	Initialize array	-	-	-	[10, 20, 30, 40, 50]
2	Access element	2	30	-	[10, 20, 30, 40, 50]
3	Update element	3	40	100	[10, 20, 30, 100, 50]
4	End	-	-	-	[10, 20, 30, 100, 50]

💡 All operations complete; array updated at index 3.

Variable Tracker

Variable	Start	After Step 2	After Step 3	Final
arr	[10, 20, 30, 40, 50]	[10, 20, 30, 40, 50]	[10, 20, 30, 100, 50]	[10, 20, 30, 100, 50]
x	undefined	30	30	30

Key Moments - 3 Insights

Why does accessing arr[2] give 30 and not 20 or 40?

Because array indexing starts at 0, so arr[2] is the third element, which is 30 as shown in step 2 of the execution table.

What happens to the array after updating arr[3] = 100?

The value at index 3 changes from 40 to 100, updating the array state as shown in step 3 of the execution table.

Can we access or update an index outside the array size?

No, accessing or updating outside the array bounds causes undefined behavior; this example only accesses valid indices as shown.

Visual Quiz - 3 Questions

Look at the execution table, what is the value of x after step 2?

A 30

B 40

C 20

D 100

💡 Hint

Check the 'Value Read' column at step 2 in the execution table.

At which step does the array change its state?

A Step 1

B Step 3

C Step 2

D Step 4

💡 Hint

Look at the 'Array State' column changes in the execution table.

If we update arr[0] = 99 instead of arr[3], what would be the array state after update?

A [10, 20, 30, 40, 50]

B [10, 20, 30, 100, 50]

C [99, 20, 30, 40, 50]

D [10, 99, 30, 40, 50]

💡 Hint

Updating index 0 changes the first element; check how array state changes in variable_tracker.

Concept Snapshot

Array Access and Update:
- Use arr[index] to read or write.
- Index starts at 0.
- Reading returns value at index.
- Writing updates value at index.
- Accessing invalid index causes error.

Full Transcript

This visual trace shows how an array is accessed and updated at specific indices. We start with an array of five integers. First, we read the value at index 2, which is 30. Then, we update the value at index 3 from 40 to 100. The variable tracker shows how the array and variable x change after each step. Key moments clarify why indexing starts at zero and how updates affect the array. The quiz tests understanding of values at each step and the effect of updates.

Practice

1. What does the expression arr[2] do in an array named arr?

easy

A. Accesses the element at index 0 of the array

B. Accesses the element at index 2 of the array

C. Updates the element at index 2 to zero

D. Deletes the element at index 2

Solution

1. Step 1: Understand array indexing

   Arrays use zero-based indexing, so index 2 means the third element.

2. Step 2: Access element at index 2

   The expression arr[2] retrieves the value stored at the third position in the array.

3. Final Answer:

   Accesses the element at index 2 of the array -> Option B

4. Quick Check:

   arr[2] means third element access [OK]

Hint: Remember arrays start at index 0, so arr[2] is third element [OK]

Common Mistakes:

• Thinking index starts at 1 instead of 0
• Confusing access with update or delete
• Assuming arr[2] changes the array

2. Which of the following is the correct syntax to update the 4th element of an integer array arr to 10 in C?

easy

A. arr[3] = 10;

B. arr(4) = 10;

C. arr[4] == 10;

D. arr{3} = 10;

Solution

1. Step 1: Identify correct array indexing syntax

   In C, arrays use square brackets with zero-based indexing, so the 4th element is at index 3.

2. Step 2: Use assignment operator correctly

   To update, use single equals =, so arr[3] = 10; is correct.

3. Final Answer:

   arr[3] = 10; -> Option A

4. Quick Check:

   Use arr[index] = value; with zero-based index [OK]

Hint: Use square brackets and single = for update, zero-based index [OK]

Common Mistakes:

• Using parentheses or braces instead of brackets
• Using double equals (==) instead of assignment (=)
• Using wrong index number for 4th element

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

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

medium

A. 6

B. 4

C. 5

D. 7

Solution

1. Step 1: Evaluate the expression arr[3] + 2

   arr[3] is the 4th element, which is 4. Adding 2 gives 6.

2. Step 2: Assign and print arr[1]

   arr[1] is updated to 6, then printed, so output is 6.

3. Final Answer:

   6 -> Option A

4. Quick Check:

   arr[3]=4 + 2 = 6 printed [OK]

Hint: Remember arr[3] is 4, add 2, assign to arr[1], print arr[1] [OK]

Common Mistakes:

• Confusing index 1 with index 3 values
• Adding incorrectly or forgetting zero-based index
• Printing wrong array element

4. Identify the error in the following C code snippet:

int arr[3] = {5, 10, 15};
arr[3] = 20;
printf("%d", arr[3]);

medium

A. Syntax error in array declaration

B. Missing semicolon after assignment

C. Array index out of bounds error

D. Incorrect printf format specifier

Solution

1. Step 1: Check array size and valid indices

   Array size is 3, valid indices are 0, 1, 2. Index 3 is out of bounds.

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

   Accessing or assigning arr[3] causes undefined behavior or runtime error.

3. Final Answer:

   Array index out of bounds error -> Option C

4. Quick Check:

   Index 3 invalid for size 3 array [OK]

Hint: Array indices go from 0 to size-1 only [OK]

Common Mistakes:

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

5. Given an integer array arr of size 6 initialized as {2, 4, 6, 8, 10, 12}, which code snippet correctly doubles the value at index 4 and then prints the entire array?

hard

A. arr[4] = arr[4] + 2;
for(int i=0; i<5; i++) printf("%d ", arr[i]);

B. arr[5] = arr[4] * 2;
for(int i=0; i<=6; i++) printf("%d ", arr[i]);

C. arr[4] *= 2;
for(int i=1; i<6; i++) printf("%d ", arr[i]);

D. arr[4] = arr[4] * 2;
for(int i=0; i<6; i++) printf("%d ", arr[i]);

Solution

1. Step 1: Correctly update value at index 4

   Index 4 is the 5th element. Doubling means multiply by 2. Both arr[4] = arr[4] * 2; and arr[4] *= 2; work, but check loop next.

2. Step 2: Print entire array correctly

   Array size is 6, so loop must run from 0 to 5 (i<6). arr[4] = arr[4] * 2; for(int i=0; i<6; i++) printf("%d ", arr[i]); uses correct loop range and prints all elements.

3. Final Answer:

   arr[4] = arr[4] * 2; for(int i=0; i<6; i++) printf("%d ", arr[i]); -> Option D

4. Quick Check:

   Update index 4, loop 0 to 5 to print all [OK]

Hint: Use zero-based index and loop from 0 to size-1 [OK]

Common Mistakes:

• Using wrong index for update
• Looping beyond array size causing errors
• Printing partial array missing elements