Diff: main.cpp

Revision:

0:733afbfa4d3f

Child:

1:fa25df63b18f

diff -r 000000000000 -r 733afbfa4d3f main.cpp
--- /dev/null	Thu Jan 01 00:00:00 1970 +0000
+++ b/main.cpp	Tue Oct 16 10:21:28 2018 +0000
@@ -0,0 +1,176 @@
+#include "mbed.h"
+#include "C12832.h"
+
+DigitalIn LDR1(p9);   //1p
+DigitalIn LDR2(p10);  //2p
+DigitalIn LDR3(p12);  //5p
+DigitalIn LDR4(p13);  //10p
+DigitalIn LDR5(p14);  //20p
+DigitalIn LDR6(p15);  //50p
+DigitalIn LDR7(p16);  //1GBP
+DigitalIn LDR8(p17);  //2GBP //I don't know the mistake 
+AnalogIn button1(p18);
+AnalogIn button2(p19);
+
+C12832 lcd(p5, p7, p6, p8, p11);
+
+/*int main()
+{
+    int j=0;
+    lcd.cls();
+    lcd.locate(0,3);
+    lcd.printf("mbed application board!");
+    
+    while(true) 
+    {   // this is the third thread
+        lcd.locate(0,15);
+        lcd.printf("Counting : %d",j);
+        j++;
+        wait(1.0);
+    }
+}*/
+
+
+int main() 
+{
+    float a[8]={0,0,0,0,0,0,0,0}; //number of 1p coins
+    int n[8]={0,0,0,0,0,0,0,0}; //amount of 1p coins
+    float v[8]={0.01,0.02,0.05,0.1,0.2,0.5,1,2};// value of coins
+    //...
+    float ta=0; //total amount of all coins
+    //int c=0; //counting variable
+    //float tm; // token money 
+    //int buttons; //one for selection of coins, one for confirming and programm ending
+    int an = 0;//number in which of the arrays we are
+    
+            time_t start, end;      //time counting
+            double elapsed;  // seconds
+            start = time(NULL);
+    
+    for(;;)
+    {
+       if(LDR1 == 1)//we could also use switch-case here for every diode
+       {
+        n[0]++;
+        a[0] = n[0]*v[0];
+        //wait(0.2); //I don't know if the programm would count a coin twice when we reset the LED immediately because the programm runs faster than the coins fall 
+        //why is LED1 = 0; a failure?
+        //c = 0;
+        time(0);
+       }
+            
+            
+   
+     end = time(NULL);
+     elapsed = difftime(end, start);
+     if (elapsed >= 8.0 /* seconds */)
+       {
+           break;
+       }
+       
+       
+       /*if(LDR2 == 1)
+       {
+        n[1]++;
+        a[1] = n[1]*v[1];
+       }
+       if(LDR3 == 1)
+       {
+        n[2]++;
+        a[2] = n[2]*v[2];
+       }
+       if(LDR4 == 1)
+       {
+        n[3]++;
+        a[3] = n[3]*v[3];
+       }
+       if(LDR5 == 1)
+       {
+        n[4]++;
+        a[4] = n[4]*v[4];
+       }
+       if(LDR6 == 1)
+       {
+        n[5]++;
+        a[5] = n[5]*v[5];
+       }
+       if(LDR7 == 1)
+       {
+        n[6]++;
+        a[6] = n[6]*v[6];
+       }
+       if(LDR8 == 1)
+       {
+        n[7]++;
+        a[7] = n[7]*v[7];
+       }*/
+       //... other types of coins
+       /*c++;
+       if (c >= 100)//depands how long the programm per cycle needs
+       {
+           break;
+       }*/ 
+    } 
+       lcd.printf("You have %d coins of 1p \nwhich means you have %f$*ps",n[0],a[0]); // it looks like we have to use only printf and not lcd.printf
+       wait(5);
+       lcd.cls();
+       //doing these things for the other coins as well
+       
+       ta = a[0]+a[1]+a[2]+a[3]+a[4]+a[5]+a[6]+a[7];
+       lcd.locate(0,0);
+       lcd.printf("Your total amount is %d %c",ta,char(163));
+       wait(5);
+       lcd.cls();
+       
+       lcd.printf("Please select witch number \nof coins you would like to see \nby pressing the button.");
+       wait(5);
+       lcd.cls();
+       lcd.printf("Please finish process than \nby using the Start-Stop button.");
+       
+       /*switch (buttons)
+       {
+           case button1:    an = an+1;
+                            if (an > 8)
+                            {
+                                an = 1
+                            }
+                            lcd.cls();
+                            lcd.printf("Do you want to see the number of %d coins?",v[an]);
+           case button2:    lcd.printf("You have %d of this type of coins. \nThat means you have %f£ of this type of coin.",n[an],a[an]); //it doesn't know button1 and button2 yet
+                            break;
+       }*/
+       while(1)
+       {
+           if (button1 == 1)
+           {
+               an++;
+               if (an > 8)
+               {
+                  an = 1;
+               }
+               lcd.cls();
+               lcd.printf("Do you want to see the number of %d coins?",v[an]);
+               //button1 = 0;
+           }
+           if (button2 == 1)
+           {
+               lcd.printf("You have %d of this type of coins. \nThat means you have %f %c of this type of coin.",n[an],a[an],char(163)); //it doesn't know button1 and button2 yet
+               break;
+           }
+       }
+       wait(5);
+       lcd.cls();
+       lcd.printf("Thank you for using 'Coounting Machine'. \nGood Bye.");
+       wait(5);
+       /*if() //press button or something like that         //total amount which is left when the user took money and entered the amount of it in the machine.
+       {
+           lcd.printf("Please enter the mount of money you took:");
+           scanf("%f",&tm);
+           lcd.printf("%f",tm);
+           wait(2);
+           ta = ta-tm;
+           lcd.printf("Now you have only %f \bx156 left", ta); //\b for one step backword and I guess that x156 for £ //I'm not sure
+       }*/
+       
+    return(0);
+}