Accessing local variables in a function.

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How can I acess the variable outside the function and thus store and show it in workspace?
function drawHexagon(sideLength, centerX, centerY)
% Calculate the coordinates of the hexagon vertices
angles = linspace(0, 2*pi, 7); % Angles for hexagon vertices
x = sideLength * sin(angles) + centerX; % x-coordinates of vertices
y = sideLength * cos(angles) + centerY; % y-coordinates of vertices
Nodes_unit_hex = [x',y'];
% Plot the hexagon
scatter(x, y );
axis equal;
grid on;
% Label the center point
text(centerX, centerY, 'C1', 'HorizontalAlignment', 'center');
end

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Star Strider
Star Strider il 29 Giu 2023
One option would be to return it as an output.
I am not certain which one you want to return, however assuming that it is ‘Nodes_unit_hex’ the function declaration would be:
function Nodes_unit_hex = drawHexagon(sideLength, centerX, centerY)
Note that whenever you call the function, it would be necessary to put a semicolon (;) at the end, to suppress printing the output even if you do not want to return it.
.
  3 Commenti
Star Strider
Star Strider il 29 Giu 2023
As always, my pleasure!
Steven Lord
Steven Lord il 29 Giu 2023
Since I've seen people make this mistake before, defining the function to have an output argument is only one part of the solution. When you call the function you need to call it with an output argument as well. MATLAB doesn't magically "poof" a variable with the same name as the output argument into the workspace from which the function was called.
Also keep in mind that the name to which you assign the output(s) of the function don't have to match the names of the output arguments in the definition, as is the case in the example below. The function declares that it returns as its first output the variable named y in its workspace. The function call assigns that data to the variable named thisIsMyOutput in its caller's workspace.
thisIsMyOutput = sample1989863(1:10)
thisIsMyOutput = 1×10
1 4 9 16 25 36 49 64 81 100
function y = sample1989863(x)
y = x.^2;
end

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Torsten
Torsten il 29 Giu 2023
Spostato: Torsten il 29 Giu 2023
By defining them as output variables
function [x,y] = drawHexagon(sideLength, centerX, centerY)
and calling the function as
[x,y] = drawHexagon(sideLength, centerX, centerY)
?

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