MATLAB: How to output optional variables in a function

function callif elsevariable output

Hi, I have a function where i want to either output a, b OR c given my input condition which is found in num.

I am getting an error. Output argument "b" (and maybe others) not assigned during call.

Can you please suggest a solution!

For example,

if num=2, i want to output a=10, b=5 and if num=4, i want to output c=100

function [a,b,c] = trial_function(num)
if num==2
a=10;
b=5 
else if num==4
c=100;
end
end

Best Answer

Whenever the user specifies a certain number of output variables on the left-hand side of an assignment, your routine needs to assign to at least that many variables starting from the left of the list.

So if the user only specifies 1 output such as if they called

v1 = trial_function(3)

then you would have to assign to at least the one variable in the left of your list, "a", and you could also assign to any of the other variables. It would, for example, be fine if you assigned to "a" and "c" but not "b" in this case; everything after the first one ("a") will be ignored.

If the user specifies two outputs, such as

[v1, v2] = trial_function(3)

then you need to assign to the first two outputs, "a" and "b", and you could also assign to any of the other variables; whatever you assign or do not assign to c will be ignored if the user only specifies two outputs. It would be an error to assign to "b" but not to "a" if the user requests two outputs.

If the user specifies

[v1, v2, v3] = trial_function(4)

then you would need to assign to all three outputs, "a", "b", and "c". The outputs v1, v2 would be given whatever value you assigned to "a" and "b": even if the user doesn't care what the values are, you must provide a value.

Please note that outside of your function, the outputs are considered only by position, not by name. If you were to do

c = trial_function(4)

then this would not match the "c" on the left side against the "c" of your output variable list: it goes strictly by position, so the outside "c" variable would be assigned the value of the variable in the first position of your list, what you call "a".

Related Solutions

MATLAB: How to solve a problem related to the output argument.

The name of output argument does not matte outside the function. If you just want to output two values for each case, just use

function [a,b] = trial_function(num)
if num==2
    a=4;
    b=5;
elseif num==4
    a=100;
    b=230;
end

Regardless, If you want to output 3rd and 4th argument in 2nd case, then you can set a and b to empty vectors

[~,~,x,y] = trial_function(4)
function [a,b,c,d] = trial_function(num)
if num==2
    a=4;
    b=5;
elseif num==4
    a = [];
    b = [];
    c=100;
    d=230;
end
end

If you want to output variable number of output arguments, then use varargout: https://www.mathworks.com/help/matlab/ref/varargout.html

MATLAB: %1, %2 and %3 in solution

You can use, for example,

maple('V3 := %3: V2 := %2: V1 := %1: subs(%3=v3,%2=v2,%1=v1,%);');

Be sure to use colon and not semi-colon until the end!

This should output a new expression that has v1, v2, v3 in place of the %1, %2, %3 subexpressions, with the values of those % subexpressions recorded in V1, V2, V3.

Note here that if you want to get V1, V2, V3 back out of Maple (because they are in Maple's engine, not MATLAB's), then you would need something like

V1 := maple('V1'); V2 := maple('V2'); V3 := maple('V3');

Note also that I assigned in reverse order. That is probably not so important, but substituting in reverse order can be important. When you substitute a variable name for a % expression then sometimes a different % expression that uses the other gets shorter enough that it is no longer long enough to break out later subexpression as a subexpression. The % expressions produced are always in increasing order, in the sense that an earlier-numbered one can refer to a later numbered one but a later numbered one can never refer to an earlier numbered one. By substituting for the last first, the nodes do not get destroyed before they would get rewritten out of existence.

Likewise if you assign from %1 first and accidentally do not suppress the assignment to %, %%, and %%% (the assignment is not done if you terminate the Maple command with colon), then when the first expression displays and builds subexpression breakouts for that, those subexpressions would get numbered starting from 1. Assigning in reverse helps reduce the impact of that (but you still need to be careful.)

I wrote above about the importance of doing the substitutions in reverse. That was a bit of a simplification of the real situation. In Maple, if you have a subs() command with multiple substitutions listed, then they are all done "simultaneously" -- e.g., the %1, %2, and %3 would be located in the expression tree before any substitution, so the order of subexpressions would not matter. But that only applies to any single subs() call, and it only applies within the Maple: I make no claim about the subs() that gets exported from the Symbolic Toolbox to the MATLAB interface.

More, perhaps, than you wanted to know ;-)

Best Answer

Related Solutions

MATLAB: How to solve a problem related to the output argument.

MATLAB: %1, %2 and %3 in solution

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