Lesson 8.5. Variable scope

Definition

The scope of a variable is the part of the code during which this variable can be used. There are 3 types of variables:

Global variables: their scope can be on several functions
Local variables : they belong to one function only
Static variables: they belong to one function only, but their value is kept.

The scope of a variable is also the area of the code where memory is allocated (reserved) for this variable:

At the declaration of the variable, the memory needed to store it is reserved.
At the end of the variable's scope (i.e. when the function or program ends), the memory is freed.

To limit the program's footprint on memory usage, we always try to reduce the scope of variables to use as little memory space as possible.

Example

Let's analyze the following example:

void fct(void);

int main(void) {
  int i=5;
  fct ();
  return 0;
}

void fct(void) {
  printf ("%d\n", i);
}

Tester l'exemple en ligne

The variable i is defined in the function main(), its scope is limited to the main program. We say that i is a local variable. It cannot be used outside of main(). Therefore, it cannot be used in the fct() function. At compile time, the following error appears:

main.c:14:17: error: use of undeclared identifier 'i'
  printf ("%d", i);
                ^
1 error generated.
compiler exit status 1

This error indicates that the variable i is not declared in the function, since it is local to the main program.

Exercises

Exercise 1

Correct the above example by adding an argument to the function fct() so that the program compiles without error. Test the program:

by calling the argument to function n: void fct (int n);
then by calling the argument to the function i: void fct (int i);

Do the exercise

Solution

Exercise 2

Write a function void increment(int n); which receives as parameter an integer n and which executes the following line: n=n+1;. Test your function with the following main program:

int main(void) {
  int n=3;
  printf ("Before, n=%d\n", n);
  increment(n);
  printf ("After, n=%d\n", n);
  return 0;
}

Do the exercise

Solution

Exercise 3

Write a function int increment(int n); which receives an integer n as a parameter and returns n+1. Test your function with the following main program:

int main(void) {
  int n=3;
  printf ("before, n=%d\n", n);
  increment(n);
  printf ("Hum... n=%d\n", n);
  n = increment(n);
  printf ("After, n=%d\n", n);
  return 0;
}

Do the exercise

Solution

Quiz

In programming, variable scope...

is the section of the code where the variable can be used

always starts with a treble clef

corresponds to the section of the code where the memory is allocated

always covers the whole program

Check

Let's consider the following function:

void fct(int i);

i is a global variable.

i is a local variable.

i is an argument of the function

The main() function must also have a variable named i.

Check

Variable scope...

must be reduced to allow an optimal management of the memory

must be constant to allow an optimal management of the memory

must be increased to allow an optimal management of the memory

None of the other answers

Check

What the following program displays:

#include <stdio.h>
void fct(n);

int main(void) {
  int n=7;
  fct(n);
  printf ("n = %d",n);
  return 0;  
}

void fct(n) {
  n=n-1;
}

n = 5

n = 6

n = 7

the program does not compile

Check Bravo! There are two distinct variables named n

What the following program displays:

#include <stdio.h>
void fct(void);

int main(void) {
  int n=7;
  fct();
  printf ("n = %d",n);
  return 0;
}

void fct(void) {
  n=n-1;
}

n = 5

n = 6

n = 7

the program does not compile

Check Bravo! The variable n is local to the main() function and unknow in the function fct().

What the following program displays:

#include <stdio.h>
int fct(int n);

int main(void) {
  int n=7;
  fct(n);
  printf ("n = %d",n);
  return 0;
}

int fct(int n) {
  return n-1;
}

n = 5

n = 6

n = 7

the program does not compile

Check Bravo! The value is returned by the function fct(), but unused in the main().

What the following program displays:

#include <stdio.h>
int fct(int n);

int main(void) {
  int n=7;
  n=fct(n);
  printf ("n = %d",n);
  return 0;
}

int fct(int n) {
  return n-1;
}

n = 5

n = 6

n = 7

the program does not compile

Check Bravo! The value is returned by the function fct() and used in the main() function.

What the following program displays:

#include <stdio.h>
int fct(int n);

int main(void) {
  int x=7;
  x=fct(x);
  printf ("x = %d",x);
  return 0;
}

int fct(int n) {
  return n-1;
}

x = 5

x = 6

x = 7

the program does not compile

Check Bravo! The value of n is assigned to x by the return keyword

What the following program displays:

#include <stdio.h>
int fct(int n);

int main(void) {
  int n;
  fct(n);
  printf ("n = %d",n);
  return 0;
}

int fct(int n) {
  return n-1;
}

n = 5

n = 6

n = 7

an arbitrary value

Check Bravo! There are two distinct variables named n. The variable in the main is not assigned.

Lesson 8.4 Summary Lesson 8.6

Lesson 8.5. Variable scope

Definition

Example

Exercises

Exercise 1

Exercise 2

Exercise 3

Quiz

See also