mark reilly
/
Counting_Machine_V5
final form
main.cpp
- Committer:
- chenniges
- Date:
- 2018-10-16
- Revision:
- 1:fa25df63b18f
- Parent:
- 0:733afbfa4d3f
- Child:
- 2:c95f08525bea
File content as of revision 1:fa25df63b18f:
#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 %c coins?",v[an],char(163)); //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); }