How do I write an if loop in a nested for loop

I am trying to create a list of instances where x is equal to y, where x and y are created by a nested for loop. However, the if statement seems to only run through the first iteration of the loop.

radiusincrement = 1;
thetaincrement = 45;
k=1;
for i=1:1:5/radiusincrement %two for loops that will sweep through wafer points
    for j=1:1:thetatotal
    r(i,j) = (i-1)*radiusincrement;
    theta(i,j) = (j-1)*thetaincrement;
    x(i,j) = r(i,j).*cos(theta(i,j).*pi/180);
    y(i,j) = r(i,j).*sin(theta(i,j).*pi/180);
    point = [x(i,j) y(i,j) 0];

    if point(1)==point(2) 
    kx(k)=x(i,j);
    ky(k)=y(i,j);
    k=k+1;
    end
    end
end

What am I doing wrong? Thanks!

Answer 1

[Rik]

Welcome to the world of floats. Just as you can't write 1/3 in decimal with a finite number of digits, computers can't store exact values sometimes. Run the code below and you will have your answer.

radiusincrement = 1;
thetaincrement = 45;
thetatotal=3;
k=1;
for i=1:1:5/radiusincrement %two for loops that will sweep through wafer points
    for j=1:1:thetatotal
         r(i,j) = (i-1)*radiusincrement;
         theta(i,j) = (j-1)*thetaincrement;
         x(i,j) = r(i,j).*cos(theta(i,j).*pi/180);
         y(i,j) = r(i,j).*sin(theta(i,j).*pi/180);
         point = [x(i,j) y(i,j) 0];
         
         if point(1)==point(2)
             kx(k)=x(i,j);
             ky(k)=y(i,j);
             k=k+1;
         elseif abs(point(1)-point(2))<=(1e-10)
             disp('you should have compared against a tolerance')
         end
    end
end

Two side notes:

If x and y are equal, why bothering storing them separately?

Did you consider doing this with an array operation?

[Walker Boldman]

haha thanks!

Yes, I considered that. The next step in this is to make a matrix with x, y, and about 4 other arrays that makes plotting and interpreting results easy. This was just getting me stuck. Thanks for the edit!

[Rik]

You're welcome. If I solved your question, please consider marking it as accepted answer, if not, feel free to comment with your remaining issues.