Revision 4:e0b82f12f958, committed 2018-10-09

Comitter:

chenniges

Date:

Tue Oct 09 10:52:39 2018 +0000

Parent:

3:2db94ee076ee

Commit message:

V2

Changed in this revision

diff -r 2db94ee076ee -r e0b82f12f958 main.cpp
--- a/main.cpp	Wed Feb 05 15:32:38 2014 +0000
+++ b/main.cpp	Tue Oct 09 10:52:39 2018 +0000
@@ -4,17 +4,69 @@
 
 C12832 lcd(p5, p7, p6, p8, p11);
 
-int main()
+/*int main()
 {
     int j=0;
     lcd.cls();
     lcd.locate(0,3);
     lcd.printf("mbed application board!");
 
-    while(true) {   // this is the third thread
+    while(true) 
+    {   // this is the third thread
         lcd.locate(0,15);
         lcd.printf("Counting : %d",j);
         j++;
         wait(1.0);
     }
+}*/
+
+
+int main() 
+{
+    int n1=0; //number of 1p coins
+    float a1=0; //amount of 1p coins
+    //...
+    float ta=0; //total amount of all coins
+    int c=0; //counting variable
+    //float tm; // token money 
+    
+  /*  for(;;)
+    {
+       if(LDR1 == 1)
+       {
+        a1++;
+        n1 = a1*0.01;
+        //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;
+       }
+       //... 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 %d$*ps",a1,n1); // 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 = n1+n2+n5+n10+n20+n50+n100+n200;
+       lcd.locate(0,0);
+       lcd.printf("Your total amount is %d",ta);
+       wait(5);
+       lcd.cls();
+       
+       /*if() //press button or something like that 
+       {
+           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);
 }