Android NDK fload addition accuracy - android

I have a problem.
I am calculating position of some points for opengl rendering.
But calculations are wrong and it results with one pixel overlap on rendered image.
And its a serious problem because it looks nasty.
Strangest thing is that error occurs with simple addition.
I am just adding 0.02 for every row and column and there is a 0,009 from nowhere.
Any ideas where its coming from?
I have tried to change compiler from arm7 to arm and it does not help
code :
for (int i=0;i<elem;i++)
{
LOGE("ROW: %d",i);
startPosX = -(float)elem*vertSize/2.0f + vertSize/2.0f;
for (int j=0;j<elem;j++)
{
LOGE("Column: %d",j);
LOGE("x: %f y: %f vs:%f",startPosX,startPosY,vertSize);
Vvertices[((i*elem*3)+j*3+0)] = startPosX;
Vvertices[((i*elem*3)+j*3+1)] = startPosY;
Vvertices[((i*elem*3)+j*3+2)] = z;
positions[i*elem+j].Set( startPosX, startPosY, z );
if(randomType)
velocity[i*elem+j].Set(getRandom()/speed,getRandom()/speed,getRandom()/speed);
else
velocity[i*elem+j].Set(0.0f,0.1f/speed,0.0f);
startPosX+=vertSize;
}
startPosY +=vertSize;
}
Column: 85
x: 0.710000 y: 0.989999 vs:0.020000
Column: 86
x: 0.729999 y: 0.989999 vs:0.020000

When you need assured precision for mathematical calculations it can sometimes be a bad idea to use types such as float, because many numbers do not have a finite representation in binary, just like some fractions do not have a finite representation in decimal.
This problem has to do with the machine representation of the numbers and not with the programming language you are using itself. Many languages with the typical ways of handling float and double types have this issue.
In Java people use classes such as BigDecimal to deal with this kind of problem.

Related

Color difference atan and atan2 difference in results

EDIT: obviously I've made ann error in atan2 after many different approaches, but still curious to find formulas.
I've read multiple topics and articles, yet do not understand why atan and atan2 give different results after convertion into degrees. Here is my example (from CIElab colour space) x -79.7751, y 2.677374209. Calculation without any changes to code will give incorrect value in both cases:
atan(-79.7751 / 2.677374209) = -1,537 / or in degrees -88,07778762
atan2(-79.7751 , 2.677374209) = 3,1258 / or in degrees 179,0965
However, that example is from an article so we can check results. Basically answer should be 271.9222. I've found over the internet correction for atan, now I've got same results, yet I still have indeterminate value for x=y=0 in atan, hence some corrections should be made for atan2. Unfortunately I did not found any that would yield same result and and would be proven (I mean what could be found in literature). Most sources for CIElab say to use atan2 without any corrections and as you can below it is not correct in my particular case. Here is what I got on this:
atan(-79.7751 / 2.677374209) = -1,537 / for atan < 0 : atan + 2pi
hence atan = 4,7459 / or in degrees 271,922 ( BINGO! )
atan2(-79.7751 , 2.677374209) = ̶3̶,̶1̶2̶5̶8̶ 3,108 / or in degrees ̶1̶7̶9̶,̶0̶9̶6̶5̶ 178,0777876
now convert, for deg > 90 : 450 - deg
atan2 result 271,922
̶A̶s̶ ̶y̶o̶u̶ ̶c̶a̶n̶ ̶s̶e̶e̶,̶ ̶̶c̶l̶o̶s̶e̶ ̶b̶u̶t̶ ̶n̶o̶ ̶c̶i̶g̶a̶r̶̶.̶ ̶A̶l̶s̶o̶ ̶i̶t̶ ̶d̶i̶f̶f̶e̶r̶s̶ ̶b̶y̶ ̶1̶ ̶d̶e̶g̶r̶e̶e̶ ̶a̶n̶d̶ ̶s̶o̶m̶e̶ ̶c̶h̶a̶n̶g̶e̶.̶ My question is how exactly I can make a correction of atan2 to get proper results?
Your formulas may be wrong, depending on the library you use. See the C++ std for atan and atan2:
atan(y/x) returns in range {-pi/2, pi/2}
atan2(y,x) returns in range {0, 2pi}
You have exchanged x,y in both cases. This has the effect of this figure:

Find movement turn angle using Location

I need find angle of vehicle turn measured in degrees.
Location points update with equal intervals (1 sec). Therefore device makes like 4-5 points during turn. I schematically displayed that on picture.
Is it possible to calculate the angle of turn using Location? If it is possible, how?
What I tried:
Create two geometric vectors from points 3, 4 and 1, 2 respectively and find angle between those vectors. Coordinates of vectors I calculated like Vector1 (lat2 - lat1; lon2 - lon2). Not sure this approach could be applied to Location coordinates.
Use location1.bearingTo(location2). But this doesn't give expected results. Seems like it gives "compass" results. Perhabs I could use it somehow but not sure.
Also tried few trigonometric formulas like here or here or here. They didn't give expected angle.
EDIT: Solution
The accepted answer works great. But to complete the answer I have to show that method of angleDifference. This one works for me:
public int getAngleDifference(int currentAngle){
int r = 0;
angleList.add(currentAngle);
if (angleList.size() == 4) {
int d = Math.abs(angleList.get(0) - angleList.get(3)) % 360;
r = d > 180 ? 360 - d : d;
angleList.clear();
}
return r;
}
I add points to list untill there're 4 of them and then calculate angle difference between 1st and 4th points for better results.
Hope it will help for someone!
vect1 = LatLon2 - LatLon1; // vector subtraction
vect2 = LatLon4 - LatLon3;
By definition of the dot product has the property:
vect1.vect2 = ||vect1||*||vect2||*Cos(theta)
Here's a breakdown of the notation
The term vect1.vect2 is the dot product of vect1 and vect2.
The general form of a dot product can be broken down component wise let v1 = <x1,y1> and v2=<x2,y2> for two arbitrary vectors v1 and v2 the dot product would be:
v1.v2 = x1*x2 + y1*y2
and the magnitude of some arbitrary vector v is:
||v|| = sqrt(v.v); which is a scalar.
The above is equivalent to the Euclidean distance formula with components x and y:
||v|| = sqrt(x^2 + y^2)
Getting the angle
Find a value for theta given the two vectors vect1 and vect2:
theta = Math.ArcCos(vect1.vect2/(||vect1||*||vect2||))
Approach 1 does not work as you described: Lat, Lon are not cartesian coordinates (One degree of longitude expressed in meters is not one degree of latitide, this is only valid at the equator). You would have first to transform to a (local) cartesian system.
An error is in the drawing: The angle marked with "?" is placed at the wrong side. You most probably want angle: 180 - ?
In your example the car ist turning less than 90°, altough your angle shows more than 90°.
To understand better make another drawing where the car turns left for only 10 degrees. In your drawing this would be 170°, which is wrong.
Approach 2) works better, but you need to sum up the angle differences.
You have to write yourself a method
double angleDifference(double angle1, double angle2);
This look easier than it is, although the code is only a few lines long.
Make sure that you have some test cases that tests the behaviour when crossing the 360° limit.
Example
(turn from bearing 10 to bearing 350), should either give 20 or -20, depending if you want that the method give sthe absolut evalue or the relative angle

Gyroscope input and Clock arthematic in UnityScript Unity3d

n00b question and I don't even understand enough to be able to Google well (or understand the answers appropriately).
I'm exploring Unity 3D and playing with Gyroscope input The issue is that each axis of the gyro returns values between -1 and +1 with 0 being the devices orientation when the gyro is enabled, where +1 & -1 share a border (like 12 on a clockface). However, the gyro's 0,0,0 point is not being reset when the gyro is enabled. The unity forums suggest deactivating the then re-activating the gyro will help - it doesn't.
In effect the gyro's 0 point is static in the current version of iOS & Unity, which means I need to 'offset' it to take into account the phone's starting position.
My plan was to have something like:
var GyroOffset : Quaternion;
var CurrentGyroAttitude : Quaternion;
var DeadSpotXPos : float;
var DeadSpotXPos : float;
var ThresholdPos : float;
var ThresholdNeg : float;
function Start() {
Input.gyro.enabled = true;
Input.gyro.UpdateInterval = 0.001;
GyroOffset = Input.gryo.attitude;
}
function Update() {
CurrentGyroAttitude = input.gyro.attitude - GyroOffset;
if (CurrentGyroAttitude.x > ThresholdPos) DoSomething(1);
if (CurrentGyroAttitude.x < ThresholdNeg) DoSomething(-1);
if (CurrentGyroAttitude.x > DeadSpotXPos &&
CurrentGyroAttitude.X < ThresholdPos) DoSomethingElse(1);
if (CurrentGyroAttitude.x < DeadSpotXNeg &&
CurrentGyroAttitude.x < ThresholdNeg) DoSomethingElse(-1);
}
* Where DoSomething(arbitraryFloat) and DoSomethingElse(anotherArbitraryFloat) does things that I want done and arbitrary float is not related to Gyro output.
As the Gyro output is between -1 and +1, we could reach a case where CurrentGyroAttitude.x could legitimately be -2 or +2. The problem with this plan is that I have to "offset" to my thresholds & deadspots as well which becomes "not neat" for tuning purposes. I'd much rather only manipulate the CurrentGyroAttitude.
However, I don't know how to do "clock face arthematic" in UnityScript. I think I'm talking about modulus mathematics but I keep running into complex mathematics about crypto that's doing my head in. :)
Can anyone walk me through "cleaning" the gyro attitude output so that my clean output stays between -1 and +1, once 'offset' has been taken into effect?
Incidentally if anyone can shed light on Unity's GyroUpdateInterval that would be awesome because the documentation doesn't explain what the updates are measured in, or what the effect of manipulating is.
I don't suppose you really wanted to access members of the Quaternion - they don't work the way you think they do. What you wanted was Quaternion.eulerAngles.
To make the angle relative to the starting position modify your Update:
Vector3 CurrentGyroEuler=(Quaternion.Inverse(Input.gyro.attitude)* GyroOffset).eulerAngles

Recognition of handwritten circles, diamonds and rectangles

I looking for some advices about recognition of three handwritten shapes - circles, diamonds and rectangles. I tried diffrent aproaches but they failed so maybe you could point me in another, better direction.
What I tried:
1) Simple algorithm based on dot product between points of handwritten shape and ideal shape. It works not so bad at recognition of rectangle, but failed on circles and diamonds. The problem is that dot product of the circle and diamond is quite similiar even for ideal shapes.
2) Same aproach but using Dynamic Time Warping as measure of simililarity. Similiar problems.
3) Neural networks. I tried few aproaches - giving points data to neural networks (Feedforward and Kohonen) or giving rasterized image. For Kohonen it allways classified all the data (event the sample used to train) into the same category. Feedforward with points was better (but on the same level as aproach 1 and 2) and with rasterized image it was very slow (I needs at least size^2 input neurons and for small sized of raster circle is indistinguishable even for me ;) ) and also without success. I think is because all of this shapes are closed figures? I am not big specialist of ANN (had 1 semester course of them) so maybe I am using them wrong?
4) Saving the shape as Freeman Chain Code and using some algorithms for computing similarity. I though that in FCC the shapes will be realy diffrent from each other. No success here (but I havent explorer this path very deeply).
I am building app for Android with this but I think the language is irrelevant here.
Here's some working code for a shape classifier. http://jsfiddle.net/R3ns3/ I pulled the threshold numbers (*Threshold variables in the code) out of the ether, so of course they can be tweaked for better results.
I use the bounding box, average point in a sub-section, angle between points, polar angle from bounding box center, and corner recognition. It can classify hand drawn rectangles, diamonds, and circles. The code records points while the mouse button is down and tries to classify when you stop drawing.
HTML
<canvas id="draw" width="300" height="300" style="position:absolute; top:0px; left:0p; margin:0; padding:0; width:300px; height:300px; border:2px solid blue;"></canvas>
JS
var state = {
width: 300,
height: 300,
pointRadius: 2,
cornerThreshold: 125,
circleThreshold: 145,
rectangleThreshold: 45,
diamondThreshold: 135,
canvas: document.getElementById("draw"),
ctx: document.getElementById("draw").getContext("2d"),
drawing: false,
points: [],
getCorners: function(angles, pts) {
var list = pts || this.points;
var corners = [];
for(var i=0; i<angles.length; i++) {
if(angles[i] <= this.cornerThreshold) {
corners.push(list[(i + 1) % list.length]);
}
}
return corners;
},
draw: function(color, pts) {
var list = pts||this.points;
this.ctx.fillStyle = color;
for(var i=0; i<list.length; i++) {
this.ctx.beginPath();
this.ctx.arc(list[i].x, list[i].y, this.pointRadius, 0, Math.PI * 2, false);
this.ctx.fill();
}
},
classify: function() {
// get bounding box
var left = this.width, right = 0,
top = this.height, bottom = 0;
for(var i=0; i<this.points.length; i++) {
var pt = this.points[i];
if(left > pt.x) left = pt.x;
if(right < pt.x) right = pt.x;
if(top > pt.y) top = pt.y;
if(bottom < pt.y) bottom = pt.y;
}
var center = {x: (left+right)/2, y: (top+bottom)/2};
this.draw("#00f", [
{x: left, y: top},
{x: right, y: top},
{x: left, y: bottom},
{x: right, y: bottom},
]);
// find average point in each sector (9 sectors)
var sects = [
{x:0,y:0,c:0},{x:0,y:0,c:0},{x:0,y:0,c:0},
{x:0,y:0,c:0},{x:0,y:0,c:0},{x:0,y:0,c:0},
{x:0,y:0,c:0},{x:0,y:0,c:0},{x:0,y:0,c:0}
];
var x3 = (right + (1/(right-left)) - left) / 3;
var y3 = (bottom + (1/(bottom-top)) - top) / 3;
for(var i=0; i<this.points.length; i++) {
var pt = this.points[i];
var sx = Math.floor((pt.x - left) / x3);
var sy = Math.floor((pt.y - top) / y3);
var idx = sy * 3 + sx;
sects[idx].x += pt.x;
sects[idx].y += pt.y;
sects[idx].c ++;
if(sx == 1 && sy == 1) {
return "UNKNOWN";
}
}
// get the significant points (clockwise)
var sigPts = [];
var clk = [0, 1, 2, 5, 8, 7, 6, 3]
for(var i=0; i<clk.length; i++) {
var pt = sects[clk[i]];
if(pt.c > 0) {
sigPts.push({x: pt.x / pt.c, y: pt.y / pt.c});
} else {
return "UNKNOWN";
}
}
this.draw("#0f0", sigPts);
// find angle between consecutive 3 points
var angles = [];
for(var i=0; i<sigPts.length; i++) {
var a = sigPts[i],
b = sigPts[(i + 1) % sigPts.length],
c = sigPts[(i + 2) % sigPts.length],
ab = Math.sqrt(Math.pow(b.x-a.x,2)+Math.pow(b.y-a.y,2)),
bc = Math.sqrt(Math.pow(b.x-c.x,2)+ Math.pow(b.y-c.y,2)),
ac = Math.sqrt(Math.pow(c.x-a.x,2)+ Math.pow(c.y-a.y,2)),
deg = Math.floor(Math.acos((bc*bc+ab*ab-ac*ac)/(2*bc*ab)) * 180 / Math.PI);
angles.push(deg);
}
console.log(angles);
var corners = this.getCorners(angles, sigPts);
// get polar angle of corners
for(var i=0; i<corners.length; i++) {
corners[i].t = Math.floor(Math.atan2(corners[i].y - center.y, corners[i].x - center.x) * 180 / Math.PI);
}
console.log(corners);
// whats the shape ?
if(corners.length <= 1) { // circle
return "CIRCLE";
} else if(corners.length == 2) { // circle || diamond
// difference of polar angles
var diff = Math.abs((corners[0].t - corners[1].t + 180) % 360 - 180);
console.log(diff);
if(diff <= this.circleThreshold) {
return "CIRCLE";
} else {
return "DIAMOND";
}
} else if(corners.length == 4) { // rectangle || diamond
// sum of polar angles of corners
var sum = Math.abs(corners[0].t + corners[1].t + corners[2].t + corners[3].t);
console.log(sum);
if(sum <= this.rectangleThreshold) {
return "RECTANGLE";
} else if(sum >= this.diamondThreshold) {
return "DIAMOND";
} else {
return "UNKNOWN";
}
} else {
alert("draw neater please");
return "UNKNOWN";
}
}
};
state.canvas.addEventListener("mousedown", (function(e) {
if(!this.drawing) {
this.ctx.clearRect(0, 0, 300, 300);
this.points = [];
this.drawing = true;
console.log("drawing start");
}
}).bind(state), false);
state.canvas.addEventListener("mouseup", (function(e) {
this.drawing = false;
console.log("drawing stop");
this.draw("#f00");
alert(this.classify());
}).bind(state), false);
state.canvas.addEventListener("mousemove", (function(e) {
if(this.drawing) {
var x = e.pageX, y = e.pageY;
this.points.push({"x": x, "y": y});
this.ctx.fillStyle = "#000";
this.ctx.fillRect(x-2, y-2, 4, 4);
}
}).bind(state), false);
Given the possible variation in handwritten inputs I would suggest that a neural network approach is the way to go; you will find it difficult or impossible to accurately model these classes by hand. LastCoder's attempt works to a degree, but it does not cope with much variation or have promise for high accuracy if worked on further - this kind of hand-engineered approach was abandoned a very long time ago.
State-of-the-art results in handwritten character classification these days is typically achieved with convolutional neural networks (CNNs). Given that you have only 3 classes the problem should be easier than digit or character classification, although from experience with the MNIST handwritten digit dataset, I expect that your circles, squares and diamonds may occasionally end up being difficult for even humans to distinguish.
So, if it were up to me I would use a CNN. I would input binary images taken from the drawing area to the first layer of the network. These may require some preprocessing. If the drawn shapes cover a very small area of the input space you may benefit from bulking them up (i.e. increasing line thickness) so as to make the shapes more invariant to small differences. It may also be beneficial to centre the shape in the image, although the pooling step might alleviate the need for this.
I would also point out that the more training data the better. One is often faced with a trade-off between increasing the size of one's dataset and improving one's model. Synthesising more examples (e.g. by skewing, rotating, shifting, stretching, etc) or spending a few hours drawing shapes may provide more of a benefit than you could get in the same time attempting to improve your model.
Good luck with your app!
A linear Hough transform of the square or the diamond ought to be easy to recognize. They will both produce four point masses. The square's will be in pairs at zero and 90 degrees with the same y-coordinates for both pairs; in other words, a rectangle. The diamond will be at two other angles corresponding to how skinny the diamond is, e.g. 45 and 135 or else 60 and 120.
For the circle you need a circular Hough transform, and it will produce a single bright point cluster in 3d (x,y,r) Hough space.
Both linear and circular Hough transforms are implemented in OpenCV, and it's possible to run OpenCV on Android. These implementations include thresholding to identify lines and circles. See pg. 329 and pg. 331 of the documentation here.
If you are not familiar with Hough transforms, the Wikipedia page is not bad.
Another algorithm you may find interesting and perhaps useful is given in this paper about polygon similarity. I implemented it many years ago, and it's still around here. If you can convert the figures to loops of vectors, this algorithm could compare them against patterns, and the similarity metric would show goodness of match. The algorithm ignores rotational orientation, so if your definition of square and diamond is with respect to the axes of the drawing surface, you will have to modify the algorithm a bit to differentiate these cases.
What you have here is a fairly standard clasification task, in an arguably vision domain.
You could do this several ways, but the best way isn't known, and can sometimes depend on fine details of the problem.
So, this isn't an answer, per se, but there is a website - Kaggle.com that runs competition for classifications. One of the sample/experiemental tasks they list is reading single hand written numeric digits. That is close enough to this problem, that the same methods are almost certainly going to apply fairly well.
I suggest you go to https://www.kaggle.com/c/digit-recognizer and look around.
But if that is too vague, I can tell you from my reading of it, and playing with that problem space, that Random Forests are a better basic starting place than Neural networks.
In this case (your 3 simple objects) you could try RanSaC-fitting for ellipse (getting the circle) and lines (getting the sides of the rectangle or diamond) - on each connected object if there are several objects to classify at the same time. Based on the actual setting (expected size, etc.) the RanSaC-parameters (how close must a point be to count as voter, how many voters you need at minimun) must be tuned. When you have found a line with RanSaC-fitting, remove the points "close" to it and go for the next line. The angles of the lines should make a distinction between diamand and rectangle easy.
A very simple approach optimized for classifying exactly these 3 objects could be the following:
compute the center of gravity of an object to classify
then compute the distances of the center to the object points as a function on the angle (from 0 to 2 pi).
classify the resulting graph based on the smoothness and/or variance and the position and height of the local maxima and minima (maybe after smoothing the graph).
I propose a way to do it in following steps : -
Take convex hull of the image (consider the shapes being convex)
divide into segments using clustering algorithms
Try to fit a curves or straight line to it and measure & threshold using training set which can be used for classifications
For your application try to divide into 4 clusters .
once you classify clusters as line or curves you can use the info to derive whether curve is circle,rectangle or diamond
I think the answers that are already in place are good, but perhaps a better way of thinking about it is that you should try to break the problem into meaningful pieces.
If possible avoid the problem entirely. For instance if you are recognizing gestures, just analyze the gestures in real time. With gestures you can provide feedback to the user as to how your program interpreted their gesture and the user will change what they are doing appropriately.
Clean up the image in question. Before you do anything come up with an algorithm to try to select what the correct thing is you are trying to analyze. Also use an appropriate filter (convolution perhaps) to remove image artifacts before you begin the process.
Once you have figured out what the thing is you are going to analyze then analyze it and return a score, one for circle, one for noise, one for line, and the last for pyramid.
Repeat this step with the next viable candidate until you come up with the best candidate that is not noise.
I suspect you will find that you don't need a complicated algorithm to find circle, line, pyramid but that it is more so about structuring your code appropriately.
If I was you I'll use already available Image Processing libraries like "AForge".
Take A look at this sample article:
http://www.aforgenet.com/articles/shape_checker
I have a jar on github that can help if you are willing to unpack it and obey the apache license. You can try to recreate it in any other language as well.
Its an edge detector. The best step from there could be to:
find the corners (median of 90 degrees)
find mean median and maximum radius
find skew/angle from horizontal
have a decision agent decide what the shape is
Play around with it and find what you want.
My jar is open to the public at this address. It is not yet production ready but can help.
Just thought I could help. If anyone wants to be a part of the project, please do.
I did this recently with identifying circles (bone centers) in medical images.
Note: Steps 1-2 are if you are grabbing from an image.
Psuedo Code Steps
Step 1. Highlight the Edges
edges = edge_map(of the source image) (using edge detector(s))
(laymens: show the lines/edges--make them searchable)
Step 2. Trace each unique edge
I would (use a nearest neighbor search 9x9 or 25x25) to identify / follow / trace each edge, collecting each point into the list (they become neighbors), and taking note of the gradient at each point.
This step produces: a set of edges.
(where one edge/curve/line = list of [point_gradient_data_structure]s
(laymens: Collect a set of points along the edge in the image)
Step 3. Analyze Each Edge('s points and gradient data)
For each edge,
if the gradient similar for a given region/set of neighbors (a run of points along an edge), then we have a straight line.
If the gradient is changing gradually, we have a curve.
Each region/run of points that is a straight line or a curve, has a mean (center) and other gradient statistics.
Step 4. Detect Objects
We can use the summary information from Step 3 to build conclusions about diamonds, circles, or squares. (i.e. 4 straight lines, that have end points near each other with proper gradients is a diamond or square. One (or more) curves with sufficient points/gradients (with a common focal point) makes a complete circle).
Note: Using an image pyramid can improve algorithm performance, both in terms of results and speed.
This technique (Steps 1-4) would get the job done for well defined shapes, and also could detect shapes that are drawn less than perfectly, and could handle slightly disconnected lines (if needed).
Note: With some machine learning techniques (mentioned by other posters), it could be helpful/important to have good "classifiers" to basically break the problem down into smaller parts/components, so then a decider further down the chain could use to better understand/"see" the objects. I think machine learning might be a little heavy-handed for this question, but still could produce reasonable results. PCA(face detection) could potentially work too.

Android calculator: manipulation query + remove unwanted zeros for the answer output

Hi I am working on an android calculator apps and the now working on the manipuations. I have defined for the following:
ArrayList<Float> inputnum = new ArrayList<Float>();
float inputnum1;
float inputnum2;
and then for the operations,
case MULTIPLY:
inputnum1 = inputnum.get(0);
inputnum2 = inputnum.get(1);
inputnum.add(inputnum1 * inputnum2);
Display.setText(String.format("%.9f", inputnum.get(0)));
similar for the division one.
The muliply function and divide function works well for integers (eg 5* 4 output 20.00000000)
however, when it deals with figures with decimal places, eg 5.3 * 4, it output as 21.12000089, which is incorrect.
what is the problem?
also, how to set output to Display to remove unnecessary zero? eg
when 5*4 it only show 20 instead of 20.000000 as final answer?
when 5.3*4 = 21.12 instead of 21.12000000 as final answer?
Thanks a lot!
Just to change all the related float to double will then avoid presenting the rounding error.
If wanted to present 9 decimal places by filling up zero after the dot, eg 7.56 become 7.560000000, can use the below coding.
Display.setText(String.format("%.9f", inputnum.get(0)));

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