Union basics - Step-by-Step Execution

Concept Flow - Union basics

Declare union variable

↓

Assign value to one member

↓

Access value via any member

↓

Value reflects last assignment

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Memory shared among members

A union lets you store different data types in the same memory space, but only one at a time. Assigning to one member changes the shared memory.

Execution Sample

C

union Data {
  int i;
  float f;
};

union Data d;
d.i = 10;

This code declares a union with int and float, creates a variable, and assigns 10 to the int member.

Execution Table

Step	Action	Member Accessed	Value Stored	Memory State (shared)
1	Declare union variable d	N/A	N/A	Uninitialized
2	Assign 10 to d.i	i	10	Memory holds int 10
3	Access d.i	i	10	Memory holds int 10
4	Access d.f	f	Bit pattern of int 10 as float	Same memory, interpreted as float

💡 Execution stops after showing union memory shared and value interpretation

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 4
d.i	undefined	10	10	10
d.f	undefined	bit pattern of 10 as float	bit pattern of 10 as float	bit pattern of 10 as float

Key Moments - 2 Insights

Why does accessing d.f after assigning d.i show a strange float value?

Because union members share the same memory, the bits of the int 10 are reinterpreted as a float, not converted. See execution_table step 4.

Can we store int and float values at the same time in a union?

No, only one member's value is stored at a time since all members share the same memory space.

Visual Quiz - 3 Questions

Look at the execution_table at step 2, what value is stored in d.i?

A Undefined

B 0

C 10

D Bit pattern of float

💡 Hint

Check the 'Value Stored' column at step 2 in execution_table

At which step does the union memory hold the int value 10?

A Step 2

B Step 4

C Step 1

D Never

💡 Hint

Look at the 'Memory State' column in execution_table

If we assign a float value to d.f after step 2, what happens to d.i?

A d.i keeps the old int value

B d.i value is overwritten by float bits

C d.i becomes the float value

D d.i and d.f store values independently

💡 Hint

Recall union shares memory; assigning to one member overwrites others

Concept Snapshot

union TypeName {
  type1 member1;
  type2 member2;
};

- All members share the same memory space.
- Only one member holds a valid value at a time.
- Assigning to one member overwrites others.
- Useful to save memory or interpret data differently.

Full Transcript

This visual trace shows how a union in C works. First, we declare a union with int and float members. Then we create a variable of that union type. When we assign 10 to the int member, the union's shared memory holds that int value. Accessing the int member returns 10 as expected. However, accessing the float member reads the same memory bits but interprets them as a float, resulting in a strange float value. This happens because all union members share the same memory space, so only one value is stored at a time. Assigning to one member overwrites the memory for all members. This helps save memory and allows different views of the same data.

Practice

1. What is the main characteristic of a union in C?

easy

A. Unions can only store integers

B. Each member has its own separate memory

C. All members share the same memory location

D. Unions automatically initialize all members

Solution

1. Step 1: Understand union memory sharing

   A union allocates one shared memory block for all its members, so they overlap.

2. Step 2: Compare with struct memory allocation

   Unlike structs, unions do not allocate separate memory for each member.

3. Final Answer:

   All members share the same memory location -> Option C

4. Quick Check:

   Union memory sharing = All members share the same memory location [OK]

Hint: Remember: union members overlap in memory [OK]

Common Mistakes:

• Thinking each member has separate memory
• Assuming unions initialize all members
• Believing unions only store one data type

2. Which of the following is the correct way to define a union named Data with an int and a float member?

easy

A. union Data { int i; float f; };

B. struct Data { int i; float f; };

C. union Data ( int i; float f; );

D. union Data = { int i; float f; };

Solution

1. Step 1: Recall union syntax

   The correct syntax uses the keyword 'union', the name, and braces with members inside.

2. Step 2: Check each option

   union Data { int i; float f; }; uses correct syntax with braces and semicolons. Options A, B, and C have syntax errors or wrong keywords.

3. Final Answer:

   union Data { int i; float f; }; -> Option A

4. Quick Check:

   Correct union syntax = union Data { int i; float f; }; [OK]

Hint: Use 'union Name { members; };' syntax [OK]

Common Mistakes:

• Using struct instead of union
• Using parentheses instead of braces
• Assigning values in the definition

3. What will be the output of the following C code?

union Number {
  int i;
  float f;
};

int main() {
  union Number num;
  num.i = 65;
  printf("%c", num.f);
  return 0;
}

medium

A. A

B. 65

C. Compilation error

D. Undefined behavior or garbage character

Solution

1. Step 1: Understand union member sharing

   Setting num.i = 65 stores integer bits in union memory.

2. Step 2: Printing num.f as char

   Printing num.f with %c treats float bits as a char, causing unpredictable output.

3. Final Answer:

   Undefined behavior or garbage character -> Option D

4. Quick Check:

   Accessing different union member types = Undefined behavior or garbage character [OK]

Hint: Accessing different union members can cause garbage output [OK]

Common Mistakes:

• Expecting integer 65 to print as 'A'
• Assuming union members convert automatically
• Ignoring type mismatch in printf

4. Identify the error in this union usage:

union Data {
  int i;
  float f;
};

int main() {
  union Data d;
  d.i = 10;
  d.f = 3.14;
  printf("%d", d.i);
  return 0;
}

medium

A. Using %d to print a union member is invalid

B. Assigning to d.f overwrites d.i value

C. Cannot assign to multiple union members sequentially

D. Union must be initialized before use

Solution

1. Step 1: Analyze assignments to union members

   Assigning d.i then d.f means d.f overwrites the same memory used by d.i.

2. Step 2: Understand printf output

   Printing d.i after assigning d.f prints the overwritten value, not 10.

3. Final Answer:

   Assigning to d.f overwrites d.i value -> Option B

4. Quick Check:

   Union members share memory, last assignment wins = Assigning to d.f overwrites d.i value [OK]

Hint: Last assigned union member overwrites previous values [OK]

Common Mistakes:

• Thinking union members hold values independently
• Believing %d cannot print union members
• Assuming union needs initialization

5. You want to store either an int, a float, or a char in the same memory space and print the stored value correctly. Which approach using a union and an additional variable is best?

hard

A. Use a union for data and an enum to track the active member type

B. Use a union and print all members every time

C. Use separate variables for each type without union

D. Use a union and always print the int member

Solution

1. Step 1: Understand union limitation

   Union stores one member at a time, so we must know which member is active.

2. Step 2: Use enum to track active member

   An enum variable can track which type is stored, so printing uses correct format.

3. Final Answer:

   Use a union for data and an enum to track the active member type -> Option A

4. Quick Check:

   Union + enum for type tracking = Use a union for data and an enum to track the active member type [OK]

Hint: Track active union member with enum for correct printing [OK]

Common Mistakes:

• Printing all union members blindly
• Not tracking which member is active
• Using separate variables wastes memory