How do I pass a dynamic number of output variables to a function?
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Some functions change their behavior dependending on the number of output variables. For example, ndgrid will create N N-Dimensional arrays where N is the number of output variables. I need to pass a number of output variables that depends on other variable (whose value is not previously known). Is it possible to do that? What is the best way to do that?
回答 (3 件)
Star Strider
2024 年 7 月 18 日 12:29
You can define as many outputs to a function as you want. In the call to the function, you can selectt specific outputs using the tilde (~) to suppress the outputs you do not need.
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Steven Lord
2024 年 7 月 18 日 14:00
I need to pass a number of output variables that depends on other variable (whose value is not previously known).
I wouldn't use the word "pass" here. Nothing from the output variables enters the workspace of the function you're calling unless those variables are also specified as input arguments.
But to answer the question about how to specify output arguments when you don't know until run-time how many there will be, use a comma-separated list. See the Function Return Values section on that page for an example that calls fileparts with three output arguments. While that example hard-codes 3 output arguments, you could create that cell array C using size information computed using a variable.
S1 = callSVD(1) % 1 output
[U3, S3, V3] = callSVD(3) % 3 outputs
In fact, you could use nargout in callSVD and avoid having to pass in the number of outputs.
S1 = callSVD2 % 1 output
[U3, S3, V3] = callSVD2 % 3 outputs
function varargout = callSVD(n)
varargout = cell(1, n);
[varargout{:}] = svd(magic(4));
end
function varargout = callSVD2 % No need for n here
varargout = cell(1, nargout); % Use nargout instead
[varargout{:}] = svd(magic(4));
end
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Walter Roberson
2024 年 7 月 18 日 17:51
OutputVariable = cell(1,NumberOfOutputsNeeded);
[OutputVariable{:}] = FunctionCall(parameters, as, appropriate);
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