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Revision 2:c95f08525bea, committed 2018-10-23
- Comitter:
- mark1998
- Date:
- Tue Oct 23 09:49:35 2018 +0000
- Parent:
- 1:fa25df63b18f
- Child:
- 3:c9057d642843
- Commit message:
- Timer working
Changed in this revision
| main.cpp | Show annotated file Show diff for this revision Revisions of this file |
--- 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.
{