Function requires more input arguments to run
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My function is:
function[fa]=speriamo(u)
if (0<=u<pi/3)
fa=1; end;
if (pi/3<=u<2*pi/3)
fa=1; end;
if (2*pi/3<=u<pi)
fa=-6*u/pi+5; end;
if (pi<=u<4*pi/3)
fa=-1; end;
if (4*pi/3<=u<5*pi/3)
fa=-1; end;
if (5*pi/3<=u<2*pi)
fa=6*u/pi-11; end;
end
when I click "Run", the program give me error: "speriamo" requires more input arguments to run. Any other variable I would add is regarded as an unnecessary variable. It's a simple function, what am I doing wrong?
4 件のコメント
Adam
2017 年 3 月 20 日
How exactly would you expect it to know what 'u' is if you just click 'Run' and don't supply it?!
Cosimo Iurlaro
2017 年 3 月 20 日
Adam
2017 年 3 月 20 日
Both of the answers below work fine to solve the problem, I am just trying to get you to think logically about it so that you don't do the same thing again.
If the whole point of your function is to run some code based on an input parameter surely the first thing you should be thinking about when running the function is to be able to pass in that input parameter, otherwise what kind of results could you expect?
Cosimo Iurlaro
2017 年 3 月 20 日
採用された回答
その他の回答 (1 件)
Steven Lord
2017 年 3 月 20 日
When you click Run you call your function with 0 input arguments. The way you have defined the function:
function[fa]=speriamo(u)
it requires one input argument to run. See this page from the documentation for how to configure the Run button to run your function with an input argument, or call your function at the MATLAB command prompt (the >> or EDU>> prompt in the Command Window) with an input argument.
FYI, this line of code:
if (0<=u<pi/3)
doesn't do what you think it does. If you're only going to run this code with a scalar (1-by-1) u then rewrite it as:
if (0 <= u) & (u < pi/3)
3 件のコメント
Cosimo Iurlaro
2017 年 3 月 20 日
jakkapally srinivas
2019 年 7 月 5 日
Do you even know what you have answered!
Steven Lord
2019 年 7 月 5 日
Yes.
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