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POa good-design approach for making a geometry library (regarding using union or not)?
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  1. POa good-design approach for making a geometry library (regarding using union or not)?
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    copied!<p>i am making a geometry library and i am confused, what should be the return type of a function which calculates the intersection of a segment with another segment. The returned value would sometimes be a point and sometimes be a segment (overlap case) and sometimes be an empty set. As per my thinking there could be 3 ways to tackle this as per follows: 1. return a union (segment, null, point) 2. return a segment with first point == second point when intersection is a single point and both points as NAN when intersection is an empty set 3. return a vector (with 0 elements for empty set, 1 element for pnt and 2 elements for segment)</p> <p>Please let me know if there are any alternates possible and what are the pros and cons of each of the design. Moreover which design is supposed to be a good design and why. I am interested in making a robust architecture which lets single pipe-lining and thus almost no rewriting of code as well as scalable (in terms of adding functionality and handling all the edge cases)</p> <p>Following is my code for reference (whose return type is vector)</p> <pre><code>vector&lt;pnt&gt; seg::inter(seg z){ vector&lt;pnt&gt; ans; if(p1==p2){if(z.doesinter(p1)){ans.pb(p1);}} else if(z.p1==z.p2){ if(doesinter(z.p1)) ans.pb(z.p1);} else{ pnt p1p2=(p2-p1); pnt q1=p1p2*pnt(0,1); long double h1,h2; if(abs((z.p2-z.p1).dot(q1))&lt;=eps){ pnt r1((z.p1-p1)/(p2-p1)),r2((z.p2-p1)/(p2-p1)); if(abs(r1.y)&lt;=eps){//colinear case h1=r1.x; h2=r2.x; if(h1&gt;h2)swap(h1,h2); if(h2&gt;=0&amp;&amp;h1&lt;=1){//add eps h1=max(0.0L,h1);h2=min(1.0L,h2); ans.pb(p1+p1p2*h1); if(doublecompare(h1,h2)==-1)ans.pb(p1+p1p2*h2);}}} else{ h1 = ((p1-z.p1).dot(q1))/((z.p2-z.p1).dot(q1)); pnt q2 = (z.p2-z.p1)*pnt(0,1); h2 = ((z.p1-p1).dot(q2))/((p2-p1).dot(q2)); if(h1+eps&gt;=0&amp;&amp;h1-eps&lt;=1&amp;&amp;h2+eps&gt;=0&amp;&amp;h2-eps&lt;=1) ans.pb(z.p1+(z.p2-z.p1)*h1);}} return ans;} </code></pre>
 

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