ishole

Determine if polyshape boundary is a hole

Syntax

TF = ishole(polyin)

TF = ishole(polyin,I)

TF = ishole(polyin) returns a logical vector whose elements are 1 (true) if the corresponding boundary of polyin is a hole.

TF = ishole(polyin,I) returns a logical vector corresponding to the boundaries of polyin indexed by I. TF is the same length as I.

Create a polygon containing one solid region and one hole, and determine which region is a hole.

t = 0.05:0.5:2*pi;
x1 = cos(t);
y1 = sin(t);
x2 = 0.5*cos(t);
y2 = 0.5*sin(t);
polyin = polyshape({x1,x2},{y1,y2})

polyin = 
  polyshape with properties:

      Vertices: [27×2 double]
    NumRegions: 1
      NumHoles: 1

plot(polyin)

Figure contains an axes object. The axes object contains an object of type polygon.

TF = ishole(polyin)

TF = 2×1 logical array

   0
   1

To query one boundary at a time, use the boundary index as a second argument.

TF = ishole(polyin,2)

TF = logical
   1

Input polyshape, specified as a scalar.

Data Types: polyshape

Boundary index, specified as a scalar integer or vector of integers. Each element of I corresponds to a single boundary of the input polyshape.

Usage notes and limitations:

Introduced in R2017b