Need help about error for Area

Need to calculate area of triangle getting error with nargchk need help to fix it.

function area = area2d(x1,y1,x2,y2,x3,y3) %function definition
msg = nargchk(1,6,nargchk);%total 6 inputs
error(msg);%error message
area=0.5*(x1(y2-y3)-x2*(y1-y3)+x3*(y1-y2)); % area calculation from the given formula

Answer 1

[Image Analyst]

Don't use area as the name of your variable. It's the name of a built in function.

Answer 2

[dpb]

You've got nargchk in the argument list as first problem...

function area = area2d(x1,y1,x2,y2,x3,y3)
  error(nargchk(6,6,nargin))
  ...

Your function as written requires 6 and only 6 arguments, not a variable number from 1 to 6...

ADDENDUM It would be "more Matlab-y" to use x- and y- arrays instead of six individual elements in which case the number of arguments would be two and you'd then check for 3 elements in each. But, this refactoring could make the upper level code much more succinct.

[RG]

im getting this error when i try to run it after the fix

function area = area2d(x1,y1,x2,y2,x3,y3) %function definition
msg = error(nargchk(6,6,nargin));%total 6 inputs
error(msg);%error message
area=0.5*(x1(y2-y3)-x2*(y1-y3)+x3*(y1-y2));

Warning: NARGCHK will be removed in a future release. Use NARGINCHK or NARGOUTCHK instead. > In area2d (line 2) Error using error Too many output arguments.

Error in area2d (line 2) msg = error(nargchk(6,6,nargin));%total 6 inputs

[dpb]

msg = error(nargchk(6,6,nargin));

Lose the assignment; error eats the message if it isn't null and throws the error...you're done at that point.

The line I coded was

error(nargchk(6,6,nargin))

altho I see I left off a closing parens; will fix that in Answer.

That's the only error-checking line you need (or want; there's no point is saving the message for anything).

[dpb]

On the warning regarding obsolescence, better practice given that would be to just write

narginchk(6,6)

and be done; it'll call error internally if needed.

Answer 3

[Steven Lord]

In addition to the narg* usage dpb mentioned, look at the first part of your last line.

area=0.5*(x1(y2-y3) ...

You don't want to try to compute element number (y2-y3) of the variable x1. You want to multiply (y2-y3) by x1.

area=0.5*(x1*(y2-y3) ...

[dpb]

Good catch, Steven, I didn't look at rest of code much (like at all).

In line with previous note on input as arrays instead of a bunch of named variables, OP may want to consider

doc polyarea

--Matlab has the functionality already supplied.

On IA's note re: use of area, it won't hurt here as it's the internal name of the return variable that goes away once the function executes so won't alias the plotting area function, but is an important point in general. If assigns the result of the function in the calling context, it would.