Wrong number of arguments to operator error

What does a “wrong number of arguments to operator” error indicate?

[Stephen23]

"What does a “wrong number of arguments to operator” error indicate?"

Clearly that wrong number of arguments have been supplied to an operator. Without knowing more about your specific situation (e.g. the operator/function/... and how you are calling it) that is the general answer to your general question. The error might even occur within some third-party code or toolbox, yet we have no idea because you have told us nothing at all about what you are doing.

[Jan]

@Scott: Please post the relevant part of the code, at least the failing line, and a copy of the complete error message. Otherwise all the readers can say is exactly, what is written in the message already.

[Scott Gronholm]

In answering a question about force in unit vector form...What is incorrect about this answer?

F=(F2+F3)i + F1 j

Where i and j are unit vectors.

[Adam]

If they are vectors then you need a multiplication '*' symbol between them and the scalars. Or do you mean they are both representing unit vectors of the complex plane, which I assume not since you are using both in the same statement?

[Stephen23]

@Scott Gronholm: what are i, j, F2, F3, and F1 exactly?

Your code might be missing a multiplication: MATLAB does not implicitly multiply two things written next to each other, which is a common shorthand people use when writing expressions. So if i is a variable then (F2+F3)i is an error, and you need (F2+F3)*i, or perhaps using element-wise multiply .*. But without knowing what those variables are this is just guessing.

[Steven Lord]

Stephen: there's at least one sort-of exception to your statement "MATLAB does not implicitly multiply two things written next to each other" and it is relevant here.

x = 2+3i

Even if there is a variable named i in the workspace, this will unambiguously represent the complex number with real part 2 and imaginary part 3. That syntax works only when what precedes the i or j is a number. It does not support variable names or expressions. These all work:

x = 2+3i
y = -4-5.3j
z = -1i

The existence of that syntax might have misled Scott into thinking his code, like example a below, would work. It doesn't. Other similar examples that will not work are below. I've listed what you should write as a comment after each incorrect command.

a = 2+(3+4)i % 2+(3+4)*1i
b = 7-pij    % 7-pi*1j
c = pi+Infi  % pi+inf*1i
d = 2+i3     % 2+3i

Of course b, c, and d won't error if you have variables (or functions that have a 0-input syntax) named pij, Infi, or i3 accessible to MATLAB. But they may not do what you expect. Note that I used 1i or 1j in the replacement expressions to avoid needing to worry if there is a variable or function named i so I could even use those inside a for loop whose loop variable is i.

for i = 1:10
  c(i) = i+2i;
end
c2 = (1:10)+2i;
isequal(c, c2) % true

Of course, defining c that way is risky. It would be easy for someone to skim the code and replace "i+2i" with "3i" or "3*i". I said I could do that, not that I should.

Answer 1

[John D'Errico]

Most operators are either monadic (one operand) or dyadic (two operands). This refers to the number of arguments the operator works on. I cannot think of any zero-adic operators off the top of my head. But I suppose you could argue i or pi to be that.

Thus transpose is a monadic operator. We can use it in the form

V'

So ' forms the transpose of V. V is the operand, and transpose works on only ONE operand.

Likewise, the unary minus is a monadic operator. Thus

-a

takes a and returns the negative of a. Note that minus has both a monadic and dyadic form, as does plus.

A dyadic operator is like the multiplication operator (* or .*). So, we can write

a.*b

Thus times takes two operands. It is dyadic.

However, we cannot write the operation

a'b

transpose is not defined as a dyadic operator. If I try to do so, I will see an error.

2'3
2'3
  ↑
Error: Unexpected MATLAB expression.

Likewise, this generates an error.

*3
*3
 ↑
Error: Unexpected MATLAB operator.

To be honest, I do not recognize the error message you have reported as occurring in any context that I can think of. So it may well be possible that some toolbox I do not have does this, even a TB written by a vendor or found on the FEX. But you have not told us what you are doing or what generated the indicated error message.

Answer 2

[John D'Errico]

Answer, based on the comment by Scott:

The thing is, it appears that i is not intended here as sqrt(-1), nor is j.

I would presume from the comment, that i is intended as a vector, typically the unit vector [1 0 0]. Likewise, j is PROBABLY intended as the unit vector [0 1 0]. These are fairly standard notations. So in that context, we would want to write

F=(F2+F3)i + F1 j

as

F = (F2+F3)*[1 0 0] + F1*[0 1 0];

Of course, you could just recognize that this reduces to the vector:

F = [(F2+F3), F1, 0];

In a strictly 2 dimensional universe, we might have

F = (F2+F3)*[1 0] + F1*[0 1];

Again, this is only conjecture. No matter what, you need to use * there to tell MATLAB to use multiplication.