Structure vs union comparison - Practice Questions

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of structure size and member values

What is the output of this C program?

C

#include <stdio.h>

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

int main() {
    struct Data s = {5, 'A', 3.14};
    printf("%d %c %.2f\n", s.i, s.c, s.d);
    printf("Size of struct Data: %zu\n", sizeof(s));
    return 0;
}

A

5 A 3.14
Size of struct Data: 12

B

5 A 3.14
Size of struct Data: 24

C

5 A 3.14
Size of struct Data: 16

D

5 A 3.14
Size of struct Data: 13

💡 Hint

Think about padding and alignment in structures.

The structure contains an int (4 bytes), a char (1 byte), and a double (8 bytes). Due to alignment, padding bytes are added so the double starts at an 8-byte boundary. This makes the total size 16 bytes on most systems.

❓ Predict Output

intermediate

Output of union member assignment

What will be printed by this C program?

C

#include <stdio.h>

union Value {
    int i;
    float f;
};

int main() {
    union Value v;
    v.i = 1065353216;
    printf("%d %.2f\n", v.i, v.f);
    return 0;
}

A 1065353216 1.00

B 1065353216 0.00

C 0 1.00

D 1065353216 1065353216.00

💡 Hint

Remember that union members share the same memory.

The integer 1065353216 has the same bit pattern as the float 1.0 in IEEE 754 format. So printing the float member shows 1.00.

🔧 Debug

advanced

Why does this union code print unexpected values?

Consider this code snippet. Why does printing both union members after assignment give unexpected results?

C

#include <stdio.h>

union Data {
    int i;
    char str[4];
};

int main() {
    union Data d;
    d.i = 0x41424344;
    printf("%d %c %c %c %c\n", d.i, d.str[0], d.str[1], d.str[2], d.str[3]);
    return 0;
}

A Because the char array is not initialized, it contains garbage values unrelated to the int.

B Because the union stores all members in the same memory, the char array shows the byte representation of the int, which depends on endianness.

C Because the int and char array have different sizes, accessing str causes a buffer overflow.

D Because the printf format specifiers are incorrect for union members.

💡 Hint

Think about how data is stored in memory and how endianness affects byte order.

Unions share the same memory for all members. Assigning to the int sets the bytes, which the char array then interprets as characters. The order depends on whether the system is little-endian or big-endian.

📝 Syntax

advanced

Which code snippet correctly declares and initializes a union?

Which of the following code snippets correctly declares a union and initializes its member?

A union Number n; n.i = 10; union Number { int i; float f; };

B union Number { int i; float f; } n; n = {10};

C union Number { int i; float f; } n = (int)10;

D union Number { int i; float f; } n = { .i = 10 };

💡 Hint

Look for correct syntax for union declaration and initialization.

Option D uses designated initializer syntax to initialize the union member correctly at declaration. Other options have syntax errors or invalid assignments.

🚀 Application

expert

Memory size comparison between struct and union

Given the following definitions, what is the total size in bytes of struct S and union U on a typical 64-bit system?

C

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

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

A struct S: 16 bytes, union U: 8 bytes

B struct S: 16 bytes, union U: 24 bytes

C struct S: 24 bytes, union U: 8 bytes

D struct S: 24 bytes, union U: 24 bytes

💡 Hint

Remember struct size includes all members plus padding; union size equals largest member size.

The struct has a char (1 byte), then padding (3 bytes), then int (4 bytes), then double (8 bytes), totaling 16 bytes due to alignment. The union size equals the largest member size, which is double (8 bytes).