mark reilly
/
Counting_Machine_V5
final form
Diff: main.cpp
- Revision:
- 2:c95f08525bea
- Parent:
- 1:fa25df63b18f
- Child:
- 3:c9057d642843
--- a/main.cpp Tue Oct 16 10:24:05 2018 +0000 +++ b/main.cpp Tue Oct 23 09:49:35 2018 +0000 @@ -42,68 +42,89 @@ //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 + Timer t; + /*time_t start, end; //time counting double elapsed; // seconds - start = time(NULL); + start = time(NULL);*/ for(;;) { + if(t > 8) + { + break; + } + if(LDR1 == 1)//we could also use switch-case here for every diode { n[0]++; a[0] = n[0]*v[0]; + t.reset(); + t.start(); //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); + //time(0); } - - + + /* end = time(NULL); elapsed = difftime(end, start); - if (elapsed >= 8.0 /* seconds */) + if (elapsed >= 8.0)//seconds { break; } - + */ - /*if(LDR2 == 1) + if(LDR2 == 1) { n[1]++; a[1] = n[1]*v[1]; + t.reset(); + t.start(); } if(LDR3 == 1) { n[2]++; a[2] = n[2]*v[2]; + t.reset(); + t.start(); } if(LDR4 == 1) { n[3]++; a[3] = n[3]*v[3]; + t.reset(); + t.start(); } if(LDR5 == 1) { n[4]++; a[4] = n[4]*v[4]; + t.reset(); + t.start(); } if(LDR6 == 1) { n[5]++; a[5] = n[5]*v[5]; + t.reset(); + t.start(); } if(LDR7 == 1) { n[6]++; a[6] = n[6]*v[6]; + t.reset(); + t.start(); } if(LDR8 == 1) { n[7]++; a[7] = n[7]*v[7]; - }*/ + t.reset(); + t.start(); + } //... other types of coins /*c++; if (c >= 100)//depands how long the programm per cycle needs @@ -111,10 +132,10 @@ break; }*/ } - lcd.printf("You have %d coins of 1p \nwhich means you have %f$*ps",n[0],a[0]); // it looks like we have to use only printf and not lcd.printf + /*lcd.printf("You have %d coins of 1p \nwhich means you have %f$*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 + //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); @@ -122,10 +143,11 @@ wait(5); lcd.cls(); - lcd.printf("Please select witch number \nof coins you would like to see \nby pressing the button."); + lcd.printf("Please select witch number \nof coins you would like to see."); wait(5); lcd.cls(); - lcd.printf("Please finish process than \nby using the Start-Stop button."); + lcd.printf("Confirm your choice by using the switch."); + wait(3); /*switch (buttons) { @@ -150,6 +172,7 @@ } lcd.cls(); lcd.printf("Do you want to see the number of %d %c coins?",v[an],char(163)); + wait(3); //button1 = 0; } if (button2 == 1) @@ -160,7 +183,7 @@ } wait(5); lcd.cls(); - lcd.printf("Thank you for using 'Coounting Machine'. \nGood Bye."); + 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. {