Extern storage class - Time & Space Complexity

Time Complexity: Extern storage class

O(n)

Understanding Time Complexity

Let's explore how using the extern storage class affects the time complexity of a program.

We want to see how the program's running time changes when accessing variables declared with extern.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

// file1.c
int count = 0;

void increment() {
    count++;
}

// file2.c
extern int count;

#include <stdio.h>

void printCount() {
    printf("Count: %d\n", count);
}
    

This code shows a variable declared in one file and accessed in another using extern.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Accessing and modifying the variable count.
• How many times: Each time increment() or printCount() is called.

How Execution Grows With Input

Accessing an extern variable is a simple operation that does not depend on input size.

Input Size (n)	Approx. Operations
10	10 variable accesses
100	100 variable accesses
1000	1000 variable accesses

Pattern observation: The number of operations grows directly with how many times the variable is accessed, but each access is constant time.

Final Time Complexity

Time Complexity: O(n)

This means the total time grows linearly with the number of times the extern variable is accessed or modified.

Common Mistake

[X] Wrong: "Using extern makes variable access slower because it is declared outside the file."

[OK] Correct: Extern just tells the compiler the variable is defined elsewhere; accessing it is just like any other variable access and takes constant time.

Interview Connect

Understanding how extern variables behave helps you reason about program structure and performance, a useful skill in many coding situations.

Self-Check

"What if the extern variable was an array and we looped over it? How would the time complexity change?"

Practice

1. What is the main purpose of the extern storage class in C?

easy

A. To declare a variable or function defined in another file

B. To allocate memory for a variable inside a function

C. To create a constant variable

D. To define a variable local to a block

Solution

1. Step 1: Understand the role of extern

   The extern keyword tells the compiler that the variable or function is defined elsewhere, usually in another file.

2. Step 2: Compare with other storage classes

   Other options like local variables or constants do not use extern. It specifically helps share variables/functions across files.

3. Final Answer:

   To declare a variable or function defined in another file -> Option A

4. Quick Check:

   extern declares external variables [OK]

Hint: Remember: extern means 'defined outside this file' [OK]

Common Mistakes:

• Confusing extern with static
• Thinking extern allocates memory
• Using extern for local variables

2. Which of the following is the correct syntax to declare an external integer variable named count?

easy

A. int extern count;

B. extern int count;

C. extern int count = 0;

D. static extern int count;

Solution

1. Step 1: Recall correct extern declaration syntax

   The correct syntax places extern before the type: extern int count;

2. Step 2: Check invalid options

   int extern count; reverses order, which is invalid. extern int count = 0; initializes the extern variable, which is not allowed. static extern int count; combines static and extern, which conflicts.

3. Final Answer:

   extern int count; -> Option B

4. Quick Check:

   extern before type, no initialization [OK]

Hint: extern must come before type and no initialization [OK]

Common Mistakes:

• Initializing extern variables
• Placing extern after type
• Combining extern with static

3. Consider two files:
// file1.c
int count = 5;
// file2.c
extern int count;
printf("%d", count);
What will be the output when these files are compiled and run together?

medium

A. 5

B. 0

C. Compilation error

D. Undefined behavior

Solution

1. Step 1: Understand extern linking

   The variable count is defined in file1.c with value 5. The extern declaration in file2.c refers to the same variable.

2. Step 2: Predict output

   Since count is defined and initialized in file1.c, printing it in file2.c outputs 5.

3. Final Answer:

   5 -> Option A

4. Quick Check:

   extern links to defined variable value [OK]

Hint: Extern links to defined variable, prints its value [OK]

Common Mistakes:

• Assuming extern creates a new variable
• Expecting zero as default
• Ignoring multi-file linking

4. What is wrong with the following code snippet?

extern int value = 10;
int main() {
  printf("%d", value);
  return 0;
}

medium

A. No variable named value declared

B. Missing semicolon after extern declaration

C. extern variable cannot be initialized

D. printf syntax is incorrect

Solution

1. Step 1: Check extern initialization rules

   An extern declaration cannot initialize the variable. Initialization must be done in exactly one place without extern.

2. Step 2: Identify error in code

   The code tries to initialize value with 10 in the extern declaration, which causes a compilation error.

3. Final Answer:

   extern variable cannot be initialized -> Option C

4. Quick Check:

   extern declarations must not initialize [OK]

Hint: Never initialize extern variables where declared [OK]

Common Mistakes:

• Initializing extern variables
• Confusing extern with normal definitions
• Ignoring linker errors

5. You have two files:
// file1.c
int total = 100;
void update() { total += 50; }
// file2.c
extern int total;
void update();
int main() {
update();
printf("%d", total);
return 0;
}
What will be the output when compiled and run together?

hard

A. Compilation error due to missing extern

B. 100

C. Undefined behavior due to multiple definitions

D. 150

Solution

1. Step 1: Understand extern and function usage

   total is defined in file1.c and declared extern in file2.c. The function update() modifies total.

2. Step 2: Trace program execution

   Main calls update(), which adds 50 to total (100 + 50 = 150). Then it prints total, so output is 150.

3. Final Answer:

   150 -> Option D

4. Quick Check:

   Extern shares variable; function updates value [OK]

Hint: Extern shares variable; function changes value before print [OK]

Common Mistakes:

• Expecting original value without update
• Forgetting to declare extern
• Confusing declaration and definition