problem with plotting the map()

Programming Questions

taar

problem with plotting the map()

in Programming Questions • 3 months ago

I read some data from arduino and after use map() to make a new range, I can't plot them:

this is my code:
what's wrong!!!!!!!!!!!!!!!

why "Hi" can not printed?

String[] a = split(trim(serial), ','); //a new array (called 'a') that stores values into separate cells (separated by commas specified in Arduino program)
float s2 = map(Integer.parseInt(a[1]),1,1023,1,400);
float s1 = map(Integer.parseInt(a[0]),1,1023,1,1023);
for (int j = 1; j <109; j++) {
if ((s1==x[j]) && (second() % 5 == 0)){
for(int i=1; i<400 ; i++){
if((s2==y[i]) && (second() % 5 == 0)){

println("Hi");
}
}
}
}

-x[j] and y[j] is two arrays to comparison the sensor value.

- best regard

Replies(4)

Chrisir

Re: problem with plotting the map()

3 months ago

in line 4 both ranges are identical

does it work without the "second"-part && (second() % 5 == 0) ?

mbraendle

Re: problem with plotting the map()

3 months ago

You are comparing two float values (s1 versus x and s2 versus y), both with an identical operator. However, the mapping transformation may result in some slight differences in the least significant digits that the values could not be equal.

Hence, if you compare floats, you should do that within a given tolerance:

s1 >= x[j] - tolerance && s1 <= x[j] + tolerance

PhiLho

Re: problem with plotting the map()

3 months ago

Alternatively, perhaps

int s2 = int(map(int(a[1]),1,1023,1,400));
int s1 = int(map(int(a[0]),1,1023,1,1023));

can work for you. (Not sure of the point of the second map() call.)

taar

Re: Re: problem with plotting the map()

3 months ago

thanks PhiLho
I rewrite my code with the changes you've proposed., The problem was related with that.

thanks again.

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