splitting decimal number to 2 pieces

Projects Programming
1

Hi!
I tried to figure this out last 2 weeks without manage to solve it(googleling, reading this forum for hours, testing different types of libraries etc). So I registered in here.
So, I try to make 3 digit 7-segment display(via three 74hc595 multiplexer) which displays decimal number for example 36.7(temperature which is always between 10.0 to 40.9 C) . To showing integer numbers is not a problem(working fine), but I can't manage to find out how to divide decimal number to two pieces (because i assume that to light decimal point on I have to have after decimal point numbers separated from number before decimal point).

Just because English isn't my native I will explain it other way too:

Problem: 37.7 (Temp which is always between 10.0 to 40.9 C )

need to get this format:

int a = 37
int b=7

to get it to the 7-segment display which will show 37.7.

-Toni
ps. Programming is new to me.

2

Demo code. You tell us

A. Why this works (work it out with paper and pen before answering.

B. How would you deal with 2 digits after the decimal point?

float temp = 37.9;
int a;
int b;

void setup() {
  // put your setup code here, to run once:
  
  a = temp; // gets the first part as going from (casting) 
  //float to int drops the bit after the decimal point
  
  Serial.begin(9600);
  Serial.print("before the point ");
  Serial.println(a);
  // get the second part
  b= temp*10-a*10;
  Serial.print("after the point ");
  Serial.print(b);
  
}

void loop() {
  // put your main code here, to run repeatedly: 
  
}

Mark

1 Like
4 
int a = (int)temp;
int b = temp - a;

No division necessary.

5

you can also use dtostrf() to create a character string out of it, and simply look for the '.' so you can pick it apart and send it, since you need a character anyway.

6

Thanks to all of you!

Answer was really really simple. But sometimes you just can't see it :slight_smile:

I solved this out this way with your help:

 const int  latchPin = 42; 
 const int  clockPin     = 41; 
 const int  dataPin    = 44; 
 
float temp=0.0;
int a=0;
int b=0;
int c;
int d;
byte g_digits[10];
byte g_digitsd[10];
float temp2=1.1;
 
 void setup()
 {
   pinMode (latchPin, OUTPUT);
   pinMode (clockPin, OUTPUT);
   pinMode (dataPin, OUTPUT);
   
   Serial.begin (115200);
   
 g_digits [0] = 1 + 2 + 4 + 8 + 16 + 32 + 00;
g_digits [1] = 0 + 2 + 4 + 0 + 00 + 00 + 00;
g_digits [2] = 1 + 2 + 0 + 8 + 16 + 00 + 64;
g_digits [3] = 1 + 2 + 4 + 8 + 00 + 00 + 64;
g_digits [4] = 0 + 2 + 4 + 0 + 00 + 32 + 64;
g_digits [5] = 1 + 0 + 4 + 8 + 00 + 32 + 64;
g_digits [6] = 1 + 0 + 4 + 8 + 16 + 32 + 64;
g_digits [7] = 1 + 2 + 4 + 0 + 00 + 00 + 00;
g_digits [8] = 1 + 2 + 4 + 8 + 16 + 32 + 64;
g_digits [9] = 1 + 2 + 4 + 8 + 00 + 32 + 64;

g_digitsd [0] = 1 + 2 + 4 + 8 + 16 + 32 + 00 +128;
g_digitsd [1] = 0 + 2 + 4 + 0 + 00 + 00 + 00 +128;
g_digitsd [2] = 1 + 2 + 0 + 8 + 16 + 00 + 64 +128;
g_digitsd [3] = 1 + 2 + 4 + 8 + 00 + 00 + 64 +128;
g_digitsd [4] = 0 + 2 + 4 + 0 + 00 + 32 + 64 +128;
g_digitsd [5] = 1 + 0 + 4 + 8 + 00 + 32 + 64 +128;
g_digitsd [6] = 1 + 0 + 4 + 8 + 16 + 32 + 64 +128;
g_digitsd [7] = 1 + 2 + 4 + 0 + 00 + 00 + 00 +128;
g_digitsd [8] = 1 + 2 + 4 + 8 + 16 + 32 + 64 +128;
g_digitsd [9] = 1 + 2 + 4 + 8 + 00 + 32 + 64 +128;

 }
 
 void loop() {
   
    temp=analogRead(A1);
    
   
  temp2=(temp/1024)*50;
 a= (int)temp2;
 b= temp2*10 - a*10;
 c=a/10;
 d=a-c*10;

Serial.println(a);
Serial.println(c);
Serial.println(d);
Serial.println(b);
Serial.println(temp2);
Serial.println("_____");


   
  
    digitalWrite(latchPin, 0);
   
    shiftOut(dataPin, clockPin, MSBFIRST, g_digits[c]);
    shiftOut(dataPin, clockPin, MSBFIRST, g_digits[d]);
    shiftOut(dataPin, clockPin, MSBFIRST, g_digitsd[b]);
   
    digitalWrite(latchPin, 1);
    delay(300);
  
  }

I just replace thermometer with variable resistor to be able to change voltage quickly. I think that there is "cleaner way to do this but it seems works so it's ok to me.

-Toni

7

Aw man... good catch. My code example tried to save the remainder to an int. You would need a float for that to work (or, as you did, multiply the result by 10 for a single digit of precision.)

Sorry about that. :slight_smile: Glad you figured it out despite my goof-up.

8

You don't need floats at all.
If you declare temp2 as an int, you can use:

// the math here is that 500/1024 equals 125/256
temp2=(temp*125)/256; // temp2 is now in tenths of a degree

only that will overflow so instead use:

// the math here is:
// 500/1024 = 125/256 = 128/256 - 3/256 = 1/2 - 3/256
temp2=(temp/2)-((3*temp)/256); // temp2 is now in tenths of a degree

Since you have the temperature in tenths of a degree, all you need to do is insert a decimal point.

9 
g_digitsd [4] = 0 + 2 + 4 + 0 + 00 + 32 + 64 +128;

Did you know about the binary constant notation?

g_digitsd [4] =  0b11100110;

Just note that the biggest bit comes first (128 +64+32+4+2) so the bits are reversed compared to your sum.