Storage size overview in C - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a variable that uses 1 byte of storage.

C

unsigned char [1] = 255;

Drag options to blanks, or click blank then click option'

Ax

Ba

Cvar1

DbyteVar

Attempts:

3 left

💡 Hint

Common Mistakes

Using int which usually uses 4 bytes.

Using variable names that are not descriptive.

The unsigned char type uses 1 byte of storage. Naming the variable byteVar is clear and descriptive.

2fill in blank

medium

Complete the code to declare a variable that uses 4 bytes of storage.

C

int [1] = 1000;

Drag options to blanks, or click blank then click option'

Avalue

Bcount

Cnum

Dtotal

Attempts:

3 left

💡 Hint

Common Mistakes

Using char which uses 1 byte.

Using variable names that are too generic.

The int type typically uses 4 bytes on modern systems. Naming the variable num is simple and clear.

3fill in blank

hard

Fix the error in the code to correctly declare a 2-byte storage variable.

C

short [1] = 32000;

Drag options to blanks, or click blank then click option'

AshortVar

Bsh

Cs

DvarShort

Attempts:

3 left

💡 Hint

Common Mistakes

Using variable names that are too short or unclear.

Confusing short with int.

The short type usually uses 2 bytes. Naming the variable shortVar clearly indicates its type.

4fill in blank

hard

Fill both blanks to declare a variable with 8 bytes storage and assign a large number.

C

[1] [2] = 9223372036854775807LL;

Drag options to blanks, or click blank then click option'

Along long

Bint

CbigNum

Dlong

Attempts:

3 left

💡 Hint

Common Mistakes

Using int which is usually 4 bytes.

Using variable names that don't describe the value.

long long is an 8-byte integer type in C. Naming the variable bigNum clearly shows it stores a big number.

5fill in blank

hard

Fill all three blanks to declare a floating-point variable with 8 bytes storage and assign a value.

C

[1] [2] = [3];

Drag options to blanks, or click blank then click option'

Adouble

Bpi

C3.141592653589793

Dfloat

Attempts:

3 left

💡 Hint

Common Mistakes

Using float which is 4 bytes.

Using unclear variable names.

double is an 8-byte floating-point type. Naming the variable pi is meaningful for this value.

Practice

1. Which of the following data types typically uses the least amount of memory in C?

easy

A. int

B. char

C. float

D. double

Solution

1. Step 1: Recall typical sizes of basic data types

   In C, char usually takes 1 byte, int 4 bytes, float 4 bytes, and double 8 bytes.

2. Step 2: Compare sizes to find the smallest

   Since 1 byte is less than 4 or 8 bytes, char uses the least memory.

3. Final Answer:

   char -> Option B

4. Quick Check:

   char = 1 byte [OK]

Hint: Remember char is 1 byte, smallest basic type in C [OK]

Common Mistakes:

• Confusing char with int size
• Assuming float is smaller than char
• Thinking double is smallest

2. Which of the following is the correct way to declare a variable of type double in C?

easy

A. double num;

B. double: num;

C. num double;

D. double-num;

Solution

1. Step 1: Recall C variable declaration syntax

   In C, the type comes first, then the variable name, ending with a semicolon.

2. Step 2: Check each option for correct syntax

   double num; follows this rule: double num;. Others have invalid syntax like colon, reversed order, or invalid characters.

3. Final Answer:

   double num; -> Option A

4. Quick Check:

   Type before name, semicolon ends declaration [OK]

Hint: Type first, then variable name, end with semicolon [OK]

Common Mistakes:

• Using colon instead of semicolon
• Reversing type and variable name
• Using invalid characters like dash

3. Consider the following C code snippet:

printf("%zu", sizeof(long));

What is the expected output on a typical 64-bit system?

medium

A. 8

B. 4

C. 2

D. 16

Solution

1. Step 1: Understand sizeof operator on 64-bit systems

   On most 64-bit systems, long is 8 bytes (64 bits).

2. Step 2: Match the output of sizeof(long)

   The sizeof operator returns the size in bytes, so it returns 8.

3. Final Answer:

   8 -> Option A

4. Quick Check:

   long = 8 bytes on 64-bit [OK]

Hint: On 64-bit, long is usually 8 bytes [OK]

Common Mistakes:

• Assuming long is 4 bytes on 64-bit
• Confusing bits with bytes
• Using wrong format specifier in printf

4. The following code tries to print the size of a short variable but causes a compilation error:

short s;
printf("%d", sizeof(s));

What is the main reason for the error?

medium

A. Variable s is not initialized

B. Missing semicolon after short s

C. Cannot use sizeof on variables

D. Using wrong format specifier for sizeof result

Solution

1. Step 1: Identify the type returned by sizeof

   The sizeof operator returns a value of type size_t, which is an unsigned integer type.

2. Step 2: Check printf format specifier compatibility

   The format specifier %d expects an int, but sizeof returns size_t. The correct specifier is %zu.

3. Final Answer:

   Using wrong format specifier for sizeof result -> Option D

4. Quick Check:

   Use %zu for sizeof output [OK]

Hint: Use %zu to print sizeof results safely [OK]

Common Mistakes:

• Using %d instead of %zu for sizeof
• Thinking sizeof needs variable initialization
• Missing semicolon (actually present)

5. You want to store 1000 integers in an array on a 32-bit system. If int is 4 bytes, how much memory in bytes will the array consume?

hard

A. 2000 bytes

B. 1000 bytes

C. 4000 bytes

D. 8000 bytes

Solution

1. Step 1: Identify size of one integer

   On a 32-bit system, int typically uses 4 bytes.

2. Step 2: Calculate total memory for 1000 integers

   Total memory = number of elements x size of each element = 1000 x 4 = 4000 bytes.

3. Final Answer:

   4000 bytes -> Option C

4. Quick Check:

   1000 x 4 = 4000 bytes [OK]

Hint: Multiply count by size of int (4 bytes) [OK]

Common Mistakes:

• Using size 2 bytes for int
• Multiplying by 8 bytes (double size)
• Confusing bits with bytes