Defining a constraint in a function

as I did not find any details about this exact case, here is my question: Is it possible to define a constraint in a function? Since my program is getting bigger, I aimed to implement functions which can be called to either define the constraint in multiple for-loops or use vectorization.

This is a minimal example of my problem

m = 16; n = 8;
A = randn(m,n);
b = randn(m,1);

x_ls = A \ b;

	cvx_solver mosek

	variable x(n)
    minimize( norm(A*x-b) )

	subject to

and the function i’m using in this case:

function constraint(x)
     x >= [1; 2; 3; 4; 5; 6; 7; 8];

This is the error message I am getting from CVX.

Incorrect number or types of inputs or outputs for function 'newcnstr'.

Error in  >=  (line 21)
b = newcnstr( evalin( 'caller', 'cvx_problem', '[]' ), x, y, '>=' );

Error in constraint (line 4)
	out = x >= [1; 2; 3; 4; 5; 6; 7; 8];

Error in test_cvx (line 17)

Maybe you can have a function return a string specifying the constraint, then eval the string in your main program. Or eval(my_constraint_function(...)) .

There are posts of mine on this forum that do some kind of string eval thing in CVX. Building Functions using CVX - #6 by Mark_L_Stone and Building Functions using CVX - #9 by Mark_L_Stone .

Also look at other posts from Search results for 'eval string' - CVX Forum: a community-driven support forum

Whether any of that is useful to you is another matter.

Thanks. The main benefit in should be to get a more reusable and compact code. Since, the CVX code in the function will be quite complicated in the end, the eval(...) option would not be very useful.

I have been able to initialize a CVX expression out(n, 1) and manually set the entries of x in the function. Afterward, the function will return the expression, and it can be used to construct a constraint in the original script.

expression out(n, 1)
constraint(out, x, n) >= (1:n).';

function f_out = constraint(in, x, n)
	for i = 1:n
		in(i) = x(i)*i;
	f_out = in;

to define a constraint like this

A = (1:n).'
x.*A >= A;

Are there any downsides to a procedure like this in CVX?

You’re in uncharted territory. You’re probably the best qualified to determine if it does what you want.