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Home/ Questions/Q 8018293
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Editorial Team
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Editorial Team
Asked: June 4, 20262026-06-04T21:05:55+00:00 2026-06-04T21:05:55+00:00

The JavaScript function function tile(u,v, a,b,c,d) { var c0 = tileCorners[a].eval(u,v); var c1 =

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The JavaScript function

function tile(u,v, a,b,c,d) {
  var c0 = tileCorners[a].eval(u,v);
  var c1 = tileCorners[b].eval(u-1,v);
  var c2 = tileCorners[c].eval(u,v-1);
  var c3 = tileCorners[d].eval(u-1,v-1);
  return c0 + c1 + c2 + c3;
}

should be equivalent to

function tile(u,v, a,b,c,d) {
  return
    tileCorners[a].eval(u,v) +
    tileCorners[b].eval(u-1,v) +
    tileCorners[c].eval(u,v-1) +
    tileCorners[d].eval(u-1,v-1);
}

yet the second function always returns undefined (the debugger will not “step into” the calls to eval) whereas the first function behaves correctly.
Is there something about having multiple eval method calls in an expression that is wrong?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-04T21:05:56+00:00Added an answer on June 4, 2026 at 9:05 pm

    You’re a victim of the rules of semicolon insertion.

    Try:

    return tileCorners[a].eval(u,v) +
    tileCorners[b].eval(u-1,v) +
    tileCorners[c].eval(u,v-1) +
    tileCorners[d].eval(u-1,v-1);
    

    Your version is equivalent to:

    return;
    
    tileCorners[a].eval(u,v) +
    tileCorners[b].eval(u-1,v) +
    tileCorners[c].eval(u,v-1) +
    tileCorners[d].eval(u-1,v-1);
    
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