Bit field structures in Embedded C - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of bit field structure size and values

What is the output of this C code when compiled and run on a typical 32-bit system?

Embedded C

typedef struct {
    unsigned int a:3;
    unsigned int b:5;
    unsigned int c:6;
} Bits;

#include <stdio.h>
int main() {
    Bits x = {5, 17, 33};
    printf("Size: %zu\n", sizeof(x));
    printf("a=%u b=%u c=%u\n", x.a, x.b, x.c);
    return 0;
}

A

Size: 4
a=5 b=17 c=1

B

Size: 4
a=5 b=17 c=33

C

Size: 2
a=5 b=17 c=1

D

Size: 2
a=5 b=17 c=33

💡 Hint

Remember that bit fields pack bits tightly and the size depends on total bits used.

The struct has 3+5+6=14 bits total, which fits in 2 bytes. The values fit in their bit widths, so c=33 is stored correctly. Size is 2 bytes on typical systems.

🧠 Conceptual

intermediate

Bit field overflow behavior

Given this bit field declaration:
unsigned int x:3;
What happens if you assign x = 10; ?

A Compilation error due to overflow

B x stores the value 2 because only the lowest 3 bits are kept

C x stores the value 10 without any change

D x stores 0 because 10 is out of range

💡 Hint

Think about how many bits 3 bits can hold and what happens to extra bits.

A 3-bit unsigned field can hold values 0 to 7. Assigning 10 (binary 1010) keeps only the lowest 3 bits (010), which is 2.

🔧 Debug

advanced

Why does this bit field struct cause unexpected output?

Consider this code snippet:

typedef struct {
    unsigned int flag:1;
    unsigned int value:7;
} Data;

int main() {
    Data d = {1, 130};
    printf("flag=%u value=%u\n", d.flag, d.value);
    return 0;
}

What is the reason for the unexpected output of value?

A The struct is not properly aligned causing garbage values

B The flag bit overlaps with value causing corruption

C The value 130 exceeds 7 bits and wraps around, causing unexpected value

D The printf format specifier is incorrect for bit fields

💡 Hint

Check the maximum value storable in 7 bits and what happens if you assign a bigger number.

7 bits can store max 127. Assigning 130 causes overflow and wraps around to 2 (130 - 128). So value prints as 2, which is unexpected.

📝 Syntax

advanced

Identify the syntax error in bit field declaration

Which option contains a syntax error in bit field declaration?

A struct S { unsigned int a:4; unsigned int b;:5 };

B struct S { unsigned int a:4; unsigned int b:5; };

C struct S { unsigned int a:4; unsigned b:5; };

D struct S { unsigned int a:4; int b:5; };

💡 Hint

Look carefully at the colon placement in bit field syntax.

Option A has a misplaced colon after 'b;' which is invalid syntax. Bit field size must follow the variable name without extra semicolons.

🚀 Application

expert

Calculate total bits and size of nested bit field structs

Given these nested structs:

typedef struct {
    unsigned int x:4;
    unsigned int y:4;
} Inner;

typedef struct {
    Inner a;
    unsigned int z:8;
} Outer;

#include 
int main() {
    printf("Size of Outer: %zu\n", sizeof(Outer));
    return 0;
}

What is the output of the program?

A Size of Outer: 4

B Size of Outer: 3

C Size of Outer: 1

D Size of Outer: 2

💡 Hint

Calculate total bits in Inner and Outer and consider alignment and padding.

Inner has 4+4=8 bits (1 byte). Outer adds 8 bits (1 byte) for z, total 16 bits (2 bytes). But due to alignment on typical 32-bit systems, sizeof(Outer) is 4 bytes.