Think like a programmer: Simplifying redundacy in C (Beginner)
Objective: Learn how to spot patterns and use them to reduce the length of your code.
Problem: Create a program that will find the Area, Radius, and Diameter of a circle when one of these values is given.
Seems simple, right?
A functional C code for this would look like the following (I wont go much into details about the source code, because that's not the point of this tutorial):
Take a close look, can you spot redundancy in the code? It's pretty obvious that we repeat too many times similar sequences of commands. There are different ways in which we could shorten and organize our code. This depends a lot on the programmer's habits, so it varies from person to person on how they organize themselves and avoid redundancy.
Take a look at this code and compare it with the previous one. You might notice that the new code is only a single line smaller than the previous code, but if you look at the size of each file, there's 0.7KB decrease in size the second code. In other words, the size of a code with barely a 100 lines decreased about 29% (without the comments, take them out if you want to see the size)! This is actually very good!
Final Notes:
From here on you should avoid redundancy if possible. Sometimes it's not worth it, for example, for small programs such as this it won't make a difference at all. However, it's something that you should take into account before you start working on large programs, otherwise you will have too much wasted memory! You can also use functions and classes to simplify things. In this case, functions would actually make the program bigger (however, more organized).
Compiled under GCC and tested in Fedora.
All the work was done by me, Riddle, please don't copy, unless you link back.
GAMEchief lessen the redundancy further:
Problem: Create a program that will find the Area, Radius, and Diameter of a circle when one of these values is given.
- Approach in reality.
First we need to know the equation we are working with. The area of the circle is described by the following equation: A = r^2 * pi
In real life, we just solve for any variable that's not given, i.e. if A = 5, then r = sqrt(A/pi) = 1.26 approx.
- Approach in a program.
Make an equation for each situation: when A is given, when D is given, and when R is given. Use if statements to move the information to the right equation.
Seems simple, right?
A functional C code for this would look like the following (I wont go much into details about the source code, because that's not the point of this tutorial):
#include <stdio.h>
#include <math.h>
int main()
{
//A = pi * r^2; A = pi * (d/2)^2
char buffer[20];
int selection = 0;
float area = 0.0;
float radius = 0.0;
float diameter = 0.0;
const float pi = 3.14159;
printf("Calculate the Area, Radius, or Diameter of a circle\n");
printf("What is the information you have?\n");
printf("1. Area\n2. Radius\n3. Diameter\n");
printf("Enter a number 1-3: ");
if(fgets(buffer, sizeof buffer, stdin) != NULL)
{
if(sscanf(buffer, "%d", &selection) == 1)
{
if(selection > 0 && selection < 4)
{
//The selection is a number between 1 and 3
if(selection == 1)
{
//We have the area, find the radius and diameter!
printf("What is the value for the area? ");
if(fgets(buffer, sizeof buffer, stdin) != NULL)
{
if(sscanf(buffer, "%f", &area) == 1)
{
//It's a valid value, lets compute!
//r = a/pi
radius = sqrt(area/pi);
diameter = 2*radius;
printf("Area = %.2f ; Radius = %.2f ; Diameter = %.2f\n", area, radius, diameter); //results up to 2 decimal digits
}
else
{
printf("That's not a valid number!\n");
}
}
}
else if(selection == 2)
{
//We have the radius, find the area and diameter!
printf("What is the value for the radius? ");
if(fgets(buffer, sizeof buffer, stdin) != NULL)
{
if(sscanf(buffer, "%f", &radius) == 1)
{
//It's a valid value, lets compute!
//A = r^2/pi
area = radius*radius*pi;
diameter = 2*(radius);
printf("Area = %.2f ; Radius = %.2f ; Diameter = %.2f\n", area, radius, diameter); //results up to 2 decimal digits
}
else
{
printf("That's not a valid number!\n");
}
}
}
else if(selection == 3)
{
//We have the diameter, find the area and radius!
printf("What is the value for the diameter? ");
if(fgets(buffer, sizeof buffer, stdin) != NULL)
{
if(sscanf(buffer, "%f", &diameter) == 1)
{
//It's a valid value, lets compute!
//A = (d/2)^2/pi
radius = diameter / 2;
area = radius*radius* pi;
printf("Area = %.2f ; Radius = %.2f ; Diameter = %.2f\n", area, radius, diameter); //results up to 2 decimal digits
}
else
{
printf("That's not a valid number!\n");
}
}
}
}
else
{
printf("No such option! BYE!\n");
}
}
}
return 0;
}
[/20]Take a close look, can you spot redundancy in the code? It's pretty obvious that we repeat too many times similar sequences of commands. There are different ways in which we could shorten and organize our code. This depends a lot on the programmer's habits, so it varies from person to person on how they organize themselves and avoid redundancy.
How Riddle would do it (this does not mean it's the best way to avoid the redundancy in this case!):
Take a look at this code and compare it with the previous one. You might notice that the new code is only a single line smaller than the previous code, but if you look at the size of each file, there's 0.7KB decrease in size the second code. In other words, the size of a code with barely a 100 lines decreased about 29% (without the comments, take them out if you want to see the size)! This is actually very good!
#include <stdio.h>
#include <math.h>
#include <string.h> //We will use strcat from here
#define PI 3.14159 //It's defined instead of declared as a const char
typedef struct //Structure to store circle data in case we want to add more circles in the future
{
float area;
float radius;
float diameter;
} Circle;
int main()
{
int selection = 0;
char a[] = "area";
char d[] = "diameter";
char r[] = "radius";
char str_f[64] = "What is the value for the "; //one of a, d, or r strings will be concatenated at the end of this string
char buffer[32]; //buffer to get the inputs
float tmp; //a temp var that will serve to hold the input for a while
//declare our structure and initiate all the values of the vars
Circle c;
c.area = 0.0;
c.radius = 0.0;
c.diameter = 0.0;
printf("Calculate the Area, Radius, or Diameter of a circle\n");
printf("What is the information you have?\n");
printf("1. Area\n2. Radius\n3. Diameter\n");
printf("Enter a number 1-3: ");
if(fgets(buffer, sizeof buffer, stdin) != NULL) //User wrote something?
{
if(sscanf(buffer, "%d", &selection) == 1)//Did they input a int?
{
if(selection > 0 && selection < 4) //is the int between 1-3?
{
if(selection == 1)//if it's 1, output the right msg about area
{
strcat(str_f,a);
printf("%s? ",str_f);
}
else if(selection == 2)
{
strcat(str_f,r);
printf("%s? ",str_f);
}
else
{
strcat(str_f,d);
printf("%s? ",str_f);
}
if(fgets(buffer, sizeof buffer, stdin) != NULL)
{
if(sscanf(buffer, "%f", &tmp))
{
if(selection == 1) //if the selection was 1, calculate the radius with the given area
{
c.area = tmp;
c.radius = sqrt(c.area/PI);
c.diameter = 2*c.radius;
}
else if(selection == 2)//calculate the area
{
c.radius = tmp;
c.area = pow(c.radius,2)*PI;
c.diameter = 2*c.radius;
}
else //calculate the diameter
{
c.diameter = tmp;
c.radius = c.diameter/2;
c.area = pow(c.radius,2)*PI;
}
printf("Area = %.2f ; Radius = %.2f ; Diameter = %.2f\n",c.area,c.radius,c.diameter);//print values
}
else
{
printf("Invalid number!\n"); //if the input is not a number print this
}
}
}
else
{
printf("No such option! BYE!\n"); //if the selection is out of range or did not input number, print this
}
}
}
return 0;//end
}
[/32][/64]Final Notes:
From here on you should avoid redundancy if possible. Sometimes it's not worth it, for example, for small programs such as this it won't make a difference at all. However, it's something that you should take into account before you start working on large programs, otherwise you will have too much wasted memory! You can also use functions and classes to simplify things. In this case, functions would actually make the program bigger (however, more organized).
Compiled under GCC and tested in Fedora.
All the work was done by me, Riddle, please don't copy, unless you link back.
GAMEchief lessen the redundancy further:
#include <stdio.h>
#include <conio.h>
#include <math.h>
#define PI 3.14159
int main()
{
char buffer[32];
char* info[3] = {"area", "diameter", "radus"};
float area = 0, diameter = 0, radius = 0, value = 0;
int selection = 0, valid = 0;
clrscr();
printf("Calculate the Area, Radius, or Diameter of a circle\nWhat is the information you have?\n1. Area\n2. Radius\n3. Diameter\nEnter a number 1-3: ");
if (fgets(buffer, sizeof buffer, stdin) != NULL && sscanf(buffer, "%d", &selection) && selection > 0 && selection < 4)
{
printf("What is the value for the %s? ", info[selection - 1]);
if (fgets(buffer, sizeof buffer, stdin) != NULL && sscanf(buffer, "%f", &value))
{
switch (selection)
{
case 1 :
area = value;
radius = sqrt(area / PI);
diameter = 2 * radius;
break;
case 2 :
radius = value;
area = PI * pow(radius, 2);
diameter = 2 * radius;
break;
default :
diameter = value;
radius = diameter / 2;
area = PI * pow(radius, 2);
}
printf("Area = %.2f\nRadius = %.2f\nDiameter = %.2f\n", area, radius, diameter);
valid = 1;
}
else
printf("Invalid number!\n");
}
else
printf("No such option!");
getch();
if (!valid)
main();
return 0;
}[/3][/32]
rk in my old ass compiler from the 90s.
