Storage size overview in C - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the size of a char in C?

A char in C is always 1 byte. It is the smallest addressable unit of memory.

beginner

How many bytes does an int typically use in C?

An int usually uses 4 bytes on modern systems, but it can vary depending on the system and compiler.

beginner

What is the difference between float and double in terms of storage size?

float typically uses 4 bytes, while double uses 8 bytes, allowing double to store more precise decimal numbers.

beginner

What does the term 'storage size' mean in C programming?

Storage size refers to how many bytes a data type or variable occupies in memory.

intermediate

Why does the size of int vary on different systems?

Because C is designed to be portable, the size of int depends on the system architecture and compiler to optimize performance.

How many bytes does a char occupy in C?

A 2 bytes

B 4 bytes

C 1 byte

D 8 bytes

A char always occupies 1 byte in C.

Which data type usually uses 8 bytes in C?

A float

B double

C int

D char

double typically uses 8 bytes to store decimal numbers with more precision.

Why can the size of int vary between systems?

A Because of different system architectures

B Because <code>int</code> is always 1 byte

C Because <code>int</code> stores characters

D Because <code>int</code> is a floating-point type

The size of int depends on system architecture and compiler design.

What does storage size measure in C?

A The number of bytes a variable occupies

B The value of a variable

C The speed of a variable

D The number of bits in a variable

Storage size measures how many bytes a variable or data type uses in memory.

Which data type is smallest in storage size?

A float

B double

C int

D char

char is the smallest data type, using 1 byte.

Explain the typical storage sizes of basic C data types like char, int, float, and double.

Think about how much memory each type usually takes on modern systems.

Why does the size of an int vary on different computers?

Consider how C adapts to different hardware.

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