Passing Parameters to Functions

PASSING PARAMETERS TO FUNCTIONS

When a function is called, the calling function may have to pass some values to the called function. We have been doing this in the programming examples given so far. We will now learn the technicalities involved in passing arguments/parameters to the called function.

There are two ways in which arguments or parameters can be passed to the called function. They include:

• Call by value in which values of variables are passed by the calling function to the called function. The programs that we have written so far call functions using call-by- value method of passing parameters.nut boltso sdt mi

• Call by reference in which address of variables are passed by the calling function to the called function.

Call by Value

Till now, we had been calling functions and passing arguments to them using call-by-value method. In the call- by-value method, the called function creates new variables to store the value of the arguments passed to it. Therefore, the called function uses a copy of the actual arguments to perform its intended task.

Programming Tip: It is legal to have multiple return statements in C.

If the called function is supposed to modify the value of the parameters passed to it, then the change will be reflected only in the called function. In the calling function no change will be made to the value of the variables. This is because all the changes were made to the copy of the variables and not to the actual variables.

To understand this concept, consider the code given below. The add() function accepts an integer variable num and adds 10 to it. In the calling function, the value of num = 2. In add (), the value of num is modified to 12 but in the calling function the change is not reflected.

#include <stdio.h>

void add(int n);

int main()

{

int num = 2;

printf("\n The value of num before calling the function = %d", num);

add (num);

printf("\n The value of num after calling the function = %d", num);

return 0;

}

void add(int n)

{

n = n + 10;

printf("\n The value of num in the called function %d", n);

}

Output

The value of num before calling the function = 2

The value of num in the called function = 12

The value of num after calling the function = 2

Since the called function uses a copy of num, the value of num in the calling function remains untouched. This concept can be more clearly understood from Figure 4.5.

In the above program, the called function could not directly modify the value of the argument that was passed to it. In case the value has to be changed, then the programmer may use the return statement. This is shown in the following code.

#include <stdio.h>

 int add (int n);

int main()

{

int num = 2;

printf("\n The value of num before calling the function %d", num);

num = add (num);

printf("\n The value of num after calling the function = %d", num);

return 0;

}

int add (int n)

{

n = n + 10;

printf("\n The value of num in the called function = %d", n);

return n;

}

Output

The value of num before calling the function = 2

The value of num in the called function = 12

The value of num after calling the function = 12

The following points are to be noted while passing arguments to a function using the call-by-value method.

• When arguments are passed by value, the called function creates new variables of the same data type as the arguments passed to it.

• The values of the arguments passed by the function are copied into the newly created variables.

• Arguments are called by value when the called function does not need to modify the values of the original variables in the calling function.

• Values of the variables in the calling function remain unaffected when the arguments are passed using call by value technique. 

Therefore, call-by-value method of passing arguments to a function must be used only in two cases:

• When the called function does not need to modify the value of the actual parameter. It simply uses the value of the parameter to perform its task.

• When you want that the called function should only temporarily modify the value of the variables and not permanently. So although the called function may modify the value of the variables, these variables remain unchanged in the calling function.

Programming Tip: Using call by value method of passing values is preferred to avoid inadvertent changes to variables of the calling function in the called function.

Pros and Cons

The biggest advantage of using the call-by-value technique to pass arguments to the called function is that arguments can be variables (e.g., x), literals (e.g., 6), or expressions (e.g., x+1).

The disadvantage is that copying data consumes additional storage space. In addition, it can take a lot of time to copy, thereby resulting in performance penalty, especially if the function is called many times.

Call by Reference

When the calling function passes arguments to the called function using call-by-value method, the only way to return the modified value of the argument to the caller is explicitly using the return statement. The better option when a function wants to modify the value of the argument is to pass arguments using call-by-reference technique. In call by reference, we declare the function parameters as references rather than normal variables. When this is done any changes made by the function to the arguments it receives are visible in the calling function.

To indicate that an argument is passed using call by reference, an asterisk (*) is placed after the type in the parameter list. This way, changes made to the parameter in the called function will then be reflected in the calling function.

Hence, in call-by-reference method, a function receives an implicit reference to the argument, rather than a copy of its value. Therefore, the function can modify the value of the variable and that change will be reflected in the calling function as well. The following program uses this concept. To understand this concept, consider the code given below.

#include <stdio.h>

void add (int *n);

int main()

{

int num = 2;

printf("\n The value of num before calling the function =  %d", num);

add (&num);

printf("\n The value of num after calling the function = %d", num);

return 0;

}

void add (int *n)

{

*n = *n + 10;

printf("\n The value of num in the called function = %d", *n);

}

Output

The value of num before calling the function = 2

The value of num in the called function = 12

The value of num after calling the function = 12

Advantages

The advantages of using the call-by-reference technique of passing arguments are as follows:

• Since arguments are not copied into new variables, it provides greater time and space efficiency.

• The called function can change the value of the argument and the change is reflected in the calling function.

• A return statement can return only one value. In case we need to return multiple values, pass those arguments by reference.

Disadvantages

However, the side-effect of using this technique is that when an argument is passed using call by address, it becomes difficult to tell whether that argument is meant for input, output, or both.

Now let us write a few programs that use both the call- by-value and the call-by-reference mechanisms.

2. Write a function to swap the value of two variables.

#include <stdio.h>

void swap_call_by_val (int, int);

void swap_call_by_ref (int *, int *);

int main()

{

int a=1, b=2, c=3, d=4;

printf("\n In main(), a = %d and b = %d", a, b);

swap_call_by_val (a, b);

printf("\n In main(), a = %d and b = %d",a, b);

printf("\n\n In main(), c = %d and d = %d", c, d);

swap_call_by_ref (&c, &d);

// address of the variables is passed

printf("\n In main(), c = %d and d = %d", c, d);

return 0;

}

Void swap_call_by_val (int a, int b)

{

int temp;

temp = a ;

a = b;

b = temp;

printf("\n In function (Call By Value Method) a = %d and b = %d", a,b);

}

void swap_call_by_ref (int *c, int *d)

{

int temp;

temp = *C;

// *operator used to refer to the value

*C = *d;

*d = temp;

printf("\n In function (Call By Reference Method) c = %d and d = %d", *c, *d);

}

Output

In main(), a = 1 and b= 2

In function (Call By Value Method) a = 2 and b = 1

In main(), a = 1 and b = 2

In main(), c = 3 and d = 4

In function (Call By Reference Method) c = 4 and d = 3

In main(), c = 4 and d = 3

3. Write a program to find biggest of three integers using functions.

#include <stdio.h>

int greater (int a, int b, int c);

int main()

{

int numl, num2, num3, large;

printf("\n Enter the first number: ");

scanf("%d", &num1);

printf("\n Enter the second number: ");

scanf("%d", &num2);

printf("\n Enter the third number: ");

scanf("%d", &num3);

large = greater (num1, num2, num3);

printf("\n Largest number %d", large);

return 0;

}

int greater (int a, int b, int c)

{

if (a>b && a>c)

return a;

if (b>a && b>c)

return b;

else

return c;

}

Output

Enter the first number: 45

Enter the second number: 23

Enter the third number: 34

Largest number = 45

4. Write a program to calculate area of a circle using function.

#include <stdio.h>

float cal area (float r);

int main()

{

float area, radius;

printf("\n Enter the radius of the circle: ");

scanf("%f", &radius);

area = cal_area (radius);

printf("\n Area of the circle with radius %f = %f", radius, area);

return 0;

}

float cal_area (float radius)

{

return (3.14 * radius * radius);

}

Output

Enter the radius of the circle: 7

Area of the circle with radius 7 = 153.83

5. Write a program to convert time to minutes.

#include <stdio.h>

#include <conio.h>

int convert_time_in_mins (int hrs, int minutes);

int main()

{

int hrs, minutes, total_mins;

printf("\n Enter hours and minutes: ");

scanf("%d %d", &hrs, &minutes);

total_mins = convert_time in mins (hrs, minutes);

printf("\n Total minutes  = %d", total_mins);

getch();

return 0;

}

int convert_time_in_mins (int hrs, int mins)

{

mins = hrs*60+ mins;

return mins;

}

Output

Enter hours and minutes: 4 30

Total minutes = 270

6. Write a program to calculate P(n/r).

#include <stdio.h>

#include <conio.h>

int Fact (int);

int main()

{

int n, r;

float result;

clrscr();

printf("\n Enter the value of n: ");

scanf("%d", &n);

printf("\n Enter the value of r: ");

scanf("%d", &r);

result = (float) Fact (n) /Fact (r);

printf("\n P(n/r): P(%d) / (%d) = %.2f", n, r, result);

getch();

return 0;

}

int Fact (int num)

{

int f=1, i;

for (i=num; i>=1; i--)

f = f*i;

return f;

}

Output

Enter the value of n: 4

Enter the value of r: 2

P(n/r): P(4) / (2) = 12.00

7. Write a program to calculate C(n/r).

#include <stdio.h>

#include <conio.h>

int Fact (int);

int main()

{

int n, r;

float result;

clrscr();

printf("\n Enter the value of n: ");

scanf("%d", &n);

printf("\n Enter the value of r: ");

scanf("%d", &r);

result = (float) Fact (n) / (Fact (r) *Fact (n-r));

printf("\n C(n/r) : C(%d/%d) = %.2f", n, r, result);

getch();

return 0;

}

int Fact (int num)

{

int f=1, i;

for (i=num; i>=1; i--)

f = f*i;

return f;

}

Output

Enter the value of n: 4

Enter the value of r: 2

C(n/r): C(4)/(2) 6.00

8. Write a program to sum the series-1/1! + 1/2! + 1/3!...........+ 1/n!.

#include <stdio.h>

#include <conio.h>

int Fact (int);

int main()

{

int n, f, i;

float result=0.0;

clrscr();

printf("\n Enter the value of n: ");

scanf("%d", &n);

for (i=1;i<=n;i++)

{

f=Fact (i);

result += 1/(float) f;

}

printf("\n The sum of the series 1/1! + 1/2! + 1/3!... = %.1f", result);

getch(),

return 0;

}

int Fact (int num)

{

int f=1, i;

for (i=num; i>=1;i--)

f = f*i;

return f;

}

Output

Enter the value of n: 2

The sum of the series 1/1! + 1/2! + 1/3!...= 1.5

9. Write a program to sum the series-1/1! + 4/2! + 27/3! +…….

#include <stdio.h>

#include <conio.h>

#include <math.h>

int Fact (int);

int main()

{

int n, i, NUM, DENO;

float sum=0.0;

clrscr();

printf("\n Enter the value of n");

scanf("%d", &n);

for (i=1;i<=n;i++)

{

NUM = pow(i,i);

DENO = Fact (i);

sum += (float) NUM/DENO;

}

printf("\n 1/1! + 4/2! + 27/3! +……%.2f", sum);

getch();

return 0;

}

int Fact (int n)

}

int f=1, i;

for (i=n;i>=1; i--)

f=f*i;

goo!

return f;

}

Output

Enter the value of n: 3

1/11 + 4/2! + 27/3! + ....= 7.50