Question

Question2 uses structured design implemented in C. Array of records (structs) with file I/O is needed.

The program takes two inputs at a time. The name of a person, and, the coin value as an integer in the range 5 to 95. Input coin values should always be divisible by 5 (integer division). Names are one word strings. An example input is:

Jane 30

This input line indicates that 30 cents change is to be given to Jane.

Output change values need to be in multiples of 50, 20, 10 and 5 cents. The program should aim to give as much of the higher valued coins as possible. A poor solution for an input of 30 cents is to give six 5 cent coins. A better solution is to give a 20 cent coin and a 10 cent coin.

Input to the program comes from a data file called coins.txt. There can be 0 and up to 10 input lines like the example above. It is also possible to have the same name repeated in the data file but the coin values can be different. When the name is the same, it would mean the same individual. If the name is the same, your program would need to add up the coin amounts to obtain a total amount for that individual before computing the change to be given.

Once your program has read in the data from coins.txt, your program will close coins.txt first, and then show a console screen menu as illustrated below. The program will continue to show the menu and execute the menu options until "Exit" is selected by entering the value 2 at the menu prompt.

1. Enter name
2. Exit

When the user enters the value 1 at the menu prompt, your program will ask for a name. As an example, if the user enters the name Jane, the program will output:

Customer:
Jane 30 cents

Change:
20 cents: 1
10 cents: 1

Change values of 0 are not shown.

If the user enters a non-existent name (e.g. Donald) at menu option 1, and therefore would not be in the array of records, your program will print:

Name: Donald

Not found

After the process output for menu option 1, the menu is redisplayed.

If the user enters 2 to exit, your program will write the coin and change data in CSV format to file called change.csv. After writing the data to the file your program will exit. In change.csv, the data line for Jane will look like the following, with each value separated by a comma and the line is terminated by newline:

Jane,30,0,1,1,0

Each data line in change.csv will be in the format:
name of person,total coin value,number of 50 cent coins, number of 20 cent coins,number of 10 cent coins,number of 5 cent coins newline.

So in the example output, Jane has 30 cents in one 20 cent coin and one 10 cent coin. There are no 50 or 5 cent coins.

The output data file change.csv cannot have repeated names.

You need to provide a test plan to fully test your algorithm and program, and provide an input data file, coins.txt, that contains test data in the specified format for testing your program.

Your solution (program and algorithm) should be modular in nature. Use a high cohesion and low coupling design. Your solution (program and algorithm) should be modular in nature.

This requires the submission of a structure chart and a high-level algorithm and suitable decompositions of each step.

Note that for this problem, the principle of code reuse is particularly important and a significant number of marks are allocated to this. You should attempt to design your solution such that it consists of a relatively small number of functions that are as general in design as possible and you should have functions/subroutines that can be reused (called repeatedly) in order to solve the majority of the problem. If you find that you have developed a large number of functions (code modules/subroutines) where each perform a similar task (or have a lot of code that is repeated in the functions) then attempt to analyse your design to generalise the logic so that you have just one general version of the function (module).

Be mindful of the cohesion exhibited by the function (module). So if you have a function (module) that is doing more than one task, then cohesion is low, and, you will need to redesign to have high cohesion.

Code to use that requires a bit of changing from question 1:

#include

coinChange(int change);
display();

int main(){

   int num = display();

   coinChange(num);

  
   return(0);
}

int coinChange(int change){
   while(change > 0){
       if(change > 0 && change <=95 && change% 5 == 0){
           if(change >= 50){
               change -= 50;
               printf("50 cents ");
           }else if(change >= 20){
               change -= 20;
               printf("20 cents ");
           }else if(change>=10){
              
               change -= 10;
               printf("10 cents ");

           }else if(change>=5){
      
               change -= 5;
               printf("5 cents ");
           }
          
       }
   }
   return change;

}

  

int display(){
   int x = 0;
   printf("Please input your coin amount: ");
   scanf("%d%*c", &x);
   return x;
}

PLEASE be mindful that everything should be in modules(methods,functions) and no global variables if they are not constants including printing out the change

Answer 1

Program:

#include<stdio.h>
#include<stdlib.h>
#include<string.h>

//structure
typedef struct
{
char name[20];
int cents;
}record;

int *coinChange(int change);

//main function
int main(){

FILE *f;

//open the file
f = fopen("coins.txt", "r");

//check if file not opened
if(f==NULL)
{
printf("File not opened!");
return 1;
}

record arr[10];
int i, j, n, op, total;
char name[20];
int *coins; // =

//read the file
for(i=0; !feof(f) ; i++)
{
fscanf(f, "%s", arr[i].name);
fscanf(f, "%d", &arr[i].cents);
}

n = i;

//close the file
fclose(f);

//loop
while(1)
{
//display menu
printf("1. Enter name\n");
printf("2. Exit\n\n");

//prompt and read option
printf("Enter option (1 or 2): ");
scanf("%d", &op);

//switch statement
switch(op)
{
case 1:
//read the name
scanf("%s", name);
total = 0;
//search in the list
for(i=0; i<n; i++)
{
if(!strcmp(name, arr[i].name))
{
total = total + arr[i].cents;
}
}
//check if nor found
if(total==0)
printf("Not found\n\n");
else
{
//get the change
coins = coinChange(total);
//print the change
if(coins[0])
printf("50 cents: %d\n", coins[0]);
if(coins[1])
printf("20 cents: %d\n", coins[1]);
if(coins[2])
printf("10 cents: %d\n", coins[2]);
if(coins[3])
printf("5 cents: %d\n", coins[3]);
printf("\n\n");
}

break;

case 2:

//create csv file
f = fopen("change.csv", "w");

//search the name in the list
for(i=0; i<n-1; i++)
{
total = arr[i].cents;
strcpy(name, arr[i].name);
for(j=i+1; j<n; j++)
{
if(!strcmp(name, arr[j].name))
{
total = total + arr[j].cents;
arr[j].cents = 0;
}
}
//check if found
if(total!=0)
{
//get change
coins = coinChange(total);
//print to the csv file
fprintf(f, "%s,%d,%d,%d,%d,%d\n", arr[i].name, total, coins[0],
coins[1],coins[2],coins[3]);
}
}

//close the file
fclose(f);
return 0;

default:
printf("Wrong option. Try again\n\n");
}
}

return(0);
}

//function to calculate the change
int* coinChange(int change){
int *coins = (int*)calloc(4, sizeof(int));

while(change > 0){
if(change > 0 && change <=95 && change% 5 == 0){
if(change >= 50){
change -= 50;
coins[0]++;
}else if(change >= 20){
change -= 20;
coins[1]++;
}else if(change>=10){
change -= 10;
coins[2]++;
}else if(change>=5){

change -= 5;
coins[3]++;
}

}
}
return coins;
}

coins.txt

Jane 20
Donald 40
Jane 10
Donald 35

Output:

1. Enter name
2. Exit

Enter option (1 or 2): 1
Jane
20 cents: 1
10 cents: 1

1. Enter name
2. Exit

Enter option (1 or 2): 2

change.csv