How to have a function return a coordinate point?
显示 更早的评论
I am trying to get a function to return (x,y) coordinates for a given angle theta. I am struggling to figure out how to get the output of the function to be coordinate points and not two seperate variables x and y.
This is what I have for the function:
function [x,y] = unitCircle(theta)
[x,y] = [cos(theta),sin(theta)]
end
And when I call the function using:
[x,y] = unitCircle(pi/4)
I get an error:
Too many output arguments.
Error in unitCircle (line 2)
[x,y] = [cos(theta),sin(theta)]
1 个评论
Voss
2023-2-12
function [x,y] = unitCircle(theta)
x = cos(theta);
y = sin(theta);
end
采纳的回答
[x,y] = unitCircle(pi/4)
function [x y] = unitCircle(theta)
z = [cos(theta),sin(theta)];
x = z(1);
y = z(2);
end
4 个评论
Connor
2023-2-12
Sulaymon Eshkabilov
2023-2-12
编辑：Sulaymon Eshkabilov
2023-2-12
There is an error in your accepted answer (proposed by Torsten). Note this code is not consistent and does not deliver (solve) for all possible values of theta. Wrong.
theta = -pi:.1:pi;
[x,y]=unitCircle(theta)
function [x y] = unitCircle(theta)
z = [cos(theta),sin(theta)];
x = z(1);
y = z(2);
end
This function file syntax has to be:
theta = -pi:.1:pi;
[x,y]=unitCircle(theta)
function [x y] = unitCircle(theta)
z = [cos(theta);sin(theta)];
x = z(1,:);
y = z(2,:);
end
Doesn't work :-)
theta = (-pi:.1:pi).';
[x,y]=unitCircle(theta)
function [x y] = unitCircle(theta)
z = [cos(theta);sin(theta)];
x = z(1,:);
y = z(2,:);
end
更多回答（1 个）
Here is the corrected code:
% Ver 1
xy = unitCircle(pi/4)
function xy = unitCircle(theta)
xy=[cos(theta); sin(theta)];
end
1 个评论
Alt. version:
% Ver 2
theta = linspace(-2*pi, 2*pi);
xy=unitCircle(theta);
plot(xy(1,:), xy(2,:)), axis equal;
xlabel("x"), ylabel("y"), grid on
function xy = unitCircle(theta)
xy(1,:)=[cos(theta)];
xy(2,:)=[sin(theta)];
end
类别
在 帮助中心 和 File Exchange 中查找有关 Logical 的更多信息
Translated by 
