Storage size overview in C - Mini Project: Build & Apply

Storage size overview

📖 Scenario: Imagine you are learning about how much space different types of data take in a computer's memory. This is important because knowing the size helps you use memory wisely in your programs.

🎯 Goal: You will write a simple C program that shows the storage size (in bytes) of different basic data types.

📋 What You'll Learn

Create variables of types char, int, float, and double

Create a variable called size_char to store the size of char

Use the sizeof operator to find sizes of the variables

Print the sizes with clear messages

💡 Why This Matters

🌍 Real World

Knowing the size of data types helps programmers write efficient code that uses memory wisely, which is important in embedded systems and performance-critical applications.

💼 Career

Understanding data sizes is a fundamental skill for software developers, especially those working in systems programming, embedded development, or performance optimization.

1

Create variables of basic data types

Create variables called c of type char, i of type int, f of type float, and d of type double.

C

#include 

int main() {
    // Create variables c, i, f, d of types char, int, float, double
    // Your code here

    return 0;
}

Hint

Use the syntax type variable_name; to declare each variable.

2

Create a variable to store size of char

Create an int variable called size_char and set it to the size of the variable c using sizeof(c).

C

#include 

int main() {
    char c;
    int i;
    float f;
    double d;

    // Create int variable size_char and assign sizeof(c)
    // Your code here

    return 0;
}

Hint

Use int size_char = sizeof(c); to get the size of c.

3

Find sizes of all variables

Create int variables called size_int, size_float, and size_double. Set each to the size of i, f, and d respectively using sizeof.

C

#include 

int main() {
    char c;
    int i;
    float f;
    double d;

    int size_char = sizeof(c);

    // Create int variables size_int, size_float, size_double
    // Assign sizes of i, f, d using sizeof
    // Your code here

    return 0;
}

Hint

Use int size_int = sizeof(i); and similarly for others.

4

Print the sizes of all variables

Use printf to print the sizes stored in size_char, size_int, size_float, and size_double. Print messages exactly as: Size of char: X bytes, replacing X with the size value.

C

#include 

int main() {
    char c;
    int i;
    float f;
    double d;

    int size_char = sizeof(c);
    int size_int = sizeof(i);
    int size_float = sizeof(f);
    int size_double = sizeof(d);

    // Print sizes with printf
    // Your code here

    return 0;
}

Hint

Use printf("Size of char: %d bytes\n", size_char); and similarly for others.

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