| CSC 161 | Grinnell College | Spring, 2015 |
| Imperative Problem Solving and Data Structures | ||
As with Scheme, C allows variables to refer to functions, and functions can be passed as parameters. However, since C requires the type of each variable or parameter be declared, variables that may refer to functions must indicate the function's signature (number and type of parameters, return type).
The remainder of this reading presents an extended example that illustrates:
Suppose you are asked to write a program to compute a circle's circumference and area for radii between 0 and 9. The desired output might be:
radius circumference area
0.0000 0.0000 0.0000
1.0000 6.2832 3.1416
2.0000 12.5664 12.5664
3.0000 18.8496 28.2743
4.0000 25.1327 50.2655
5.0000 31.4159 78.5398
6.0000 37.6991 113.0973
7.0000 43.9823 153.9380
8.0000 50.2655 201.0619
9.0000 56.5487 254.4690
One simple approach for writing this type of program might utilize these features:
The resulting C program follows:
/* program to compute a circle's circumference and area
*/
# include <stdio.h>
const double pi = 3.1415926535;
/* circumference function */
double circum (double radius)
{
return 2 * pi * radius;
}
/* area function */
double area (double radius)
{
return pi * radius * radius;
}
int main ()
{
printf (" radius circumference area\n");
double radius;
for (radius = 0; radius < 10; radius++)
{
printf ("%12.4lf", radius);
printf ("%12.4lf", circum(radius));
printf ("%12.4lf", area(radius));
printf ("\n");
}
return 0;
}
The above C program works fine and likely is completely satisfactory for the simple problem given. However, several elements in the program contain some common elements. For example,
Although these common elements are hardly earth shaking, it can be helpful to take advantage of such common elements in more complicated programs.
In exploiting the similarities in the printf statements, we might write a function that takes a radius and a function as parameters and then performs the required printing. The following code is an example:
/* printing function */
double myPrint (double x, double f (double))
{
printf ("%12.4lf", f(x));
}
This function has two formal parameters, the number x and a function f. Further, f is identified as a function that will utilize one double value as an input parameter, and f will return a double when it completes execution.
The function myPrint utilizes the function f when it performs its printing.
The following program replaces the printf statements in the main loop by calls to myPrint.
Note that myPrint is also used for printing the value of radius by using an identify function. The code observes, however, that a simple printf for radius could be used as a reasonable alternative.
/* program to compute a circle's circumference and area
example using function with a function parameter
*/
# include <stdio.h>
const double pi = 3.1415926535;
/* identity function */
double iden (double radius)
{
return radius;
}
/* circumference function */
double circum (double radius)
{
return 2 * pi * radius;
}
/* area function */
double area (double radius)
{
return pi * radius * radius;
}
/* printing function */
void myPrint (double x, double f (double))
{
printf ("%12.4lf", f(x));
}
int main ()
{
printf (" radius circumference area\n");
double radius;
for (radius = 0; radius < 10; radius++)
{
myPrint (radius, iden);
/* could also be
printf ("%12.4lf", radius);
*/
myPrint (radius, circum);
myPrint (radius, area);
printf ("\n");
}
return 0;
}
As noted earlier in this lab, functions as parameters provide one mechanism to take advantage of common elements within an algorithm. A second approach involves utilizing an array of functions.
Already, we have observed that the main loop in our example utilizes the same printf statement — with different functions being called. This suggests that we might declare an array of the relevant functions:
double (*funcarr[3]) (double)= {iden, circum, area};
Let's unpack this syntax:
We can declare a simple variable as a pointer to a function with the statement:
double (*f) (double)
This states that f will refer to a function that has one parameter (a double) and it will return a double. (The * means that f will be a function pointer — a reference to a function.)
With this declaration, f can be assigned a function:
f = area;
so the statement
printf ("%5.2", f(3.5));
will cause the area function to be called, executed with the value 3.5, and used as the value to be printed.
As with many C declarations, a variable can be defined and initialized in a single statement. The following declaration initializes f as a reference to the area function.
double (*f) (double) = area;
Arrays are declared by placing [] within the declaration:
double (*funcarr[3]) (double);
Here funcarr is declared as an array of 3 function pointers.
double (*funcarr[3]) (double)= {iden, circum, area};
With this declaration, we can refer to the circum function as
funcarr[1]
and we can call this function with the parameter 3.5 by writing
funcarr[1] (3.5)
The following program uses this declaration of funcarr in the main loop by iterating through the three functions of interest for printing.
/* program to compute a circle's circumference and area
example using an array of functions
*/
# include <stdio.h>
const double pi = 3.1415926535;
/* identity function */
double iden (double radius)
{
return radius;
}
/* circumference function */
double circum (double radius)
{
return 2 * pi * radius;
}
/* area function */
double area (double radius)
{
return pi * radius * radius;
}
int main ()
{
double (*funcarr[3]) (double) = {iden, circum, area};
printf (" radius circumference area\n");
double radius;
for (radius = 0; radius < 10; radius++)
{
int i;
for (i = 0; i < 3; i++)
{
printf ("%12.4lf", funcarr[i](radius));
}
printf ("\n");
}
return 0;
}
Development of laboratory exercises is an iterative process. Prof. Walker welcomes your feedback! Feel free to talk to him during class or stop by his office.