Storage size overview in C - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this code snippet?

Consider the following C code that prints the size of different data types using sizeof. What will be the output?

C

#include <stdio.h>

int main() {
    printf("%zu %zu %zu\n", sizeof(char), sizeof(int), sizeof(double));
    return 0;
}

A 1 2 4

B 1 4 8

C 2 4 8

D 1 8 8

💡 Hint

Remember that char is usually 1 byte, int is commonly 4 bytes on modern systems, and double is often 8 bytes.

The sizeof operator returns the size in bytes of the given type. On most modern systems, char is 1 byte, int is 4 bytes, and double is 8 bytes.

❓ Predict Output

intermediate

What is the size of a pointer on a 64-bit system?

Given the following code, what will be the output on a typical 64-bit system?

C

#include <stdio.h>

int main() {
    int *p;
    printf("%zu\n", sizeof(p));
    return 0;
}

A 8

B 16

C 2

D 4

💡 Hint

Think about the size of pointers on 64-bit architectures.

On 64-bit systems, pointers typically occupy 8 bytes because they need to store 64-bit addresses.

❓ Predict Output

advanced

What is the output of this struct size code?

Consider this C code that prints the size of a struct with different data types. What is the output?

C

#include <stdio.h>

struct Data {
    char c;
    int i;
    double d;
};

int main() {
    printf("%zu\n", sizeof(struct Data));
    return 0;
}

A 16

B 20

C 13

D 24

💡 Hint

Remember that structs may have padding to align data properly in memory.

The struct contains a char (1 byte), followed by 3 bytes padding, an int (4 bytes), and a double (8 bytes). Due to padding for alignment, the total size is 24 bytes on typical systems.

❓ Predict Output

advanced

What is the output of this union size code?

What will this program print when it runs?

C

#include <stdio.h>

union U {
    char c;
    int i;
    double d;
};

int main() {
    printf("%zu\n", sizeof(union U));
    return 0;
}

A 16

B 4

C 8

D 1

💡 Hint

Unions allocate memory equal to the size of their largest member, possibly with padding.

The largest member is double which is 8 bytes. The union size is 8 bytes on typical systems.

🧠 Conceptual

expert

How many bytes does this bit-field struct occupy?

Given the following struct with bit-fields, how many bytes does it occupy on a typical 64-bit system?

C

struct BitField {
    unsigned int a : 3;
    unsigned int b : 5;
    unsigned int c : 6;
    unsigned int d : 18;
};

A 4

B 8

C 2

D 6

💡 Hint

Bit-fields are packed into the smallest number of bytes that can hold all bits, aligned to the base type size.

The total bits are 3 + 5 + 6 + 18 = 32 bits, which fits exactly into 4 bytes (size of unsigned int). So the struct occupies 4 bytes.

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