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About TheUnbeliever

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  1. TheUnbeliever

    Hundreds of "Identifier not found" errors

    Yes, that's Yes, it would - that's why two separate people have told you so :-) You may have similar problems in other source files, but that is the cause of all the errors you've posted here. Roughly what's happening here is: 1. Compiler looks at function definition. 2. Compiler sees that it claims to be a member function, but there is no matching declaration. 3. Compiler tells you so. 4. Compiler assumes that it is not a member function, and proceeds to see what sense it can make of the rest. 5. Because this assumption is wrong, Compiler thinks things which you expect to be in scope (e.g. timer) are not in scope. 6. Compiler tells you so, repeatedly. You have to take errors after the first with a pinch of salt. With practice, you'll learn to tell when an earlier error has caused the others. Until then, as I say, a reasonable rule of thumb is that if you get squillions of errors, then one of the first few is the real problem. I've had GCC spit 40 pages of errors at me, all to be fixed by changing just a couple characters in some header somewhere.
  2. TheUnbeliever

    Hundreds of "Identifier not found" errors

    Well, for example, the 'Act_Shoot' declaration in FINALBOSS.h is commented out. The compiler complains about that, and will then spew out a whole collection of errors for the body of the function because e.g. as it's not a member function, it doesn't have (unqualified) access to other members of an instance like 'timer'. More generally, a small number of errors is often unrelated, but a huge number suggests that there's a few root causes cascading, and that you're better taking the first, recompiling and repeating the process until they're hacked down to a more sensible number.
  3. You can index a linear array as if it were multidimensional by using a formula like x + y * width. i.e. Skip y rows of width elements, then x elements preceding the one you want on the row you want. Your code just reverses this. Where width is the number of tiles in the texture, and x < width: integer division truncates so (x + y * width) / width = y; % is the modulo operation, which effectively gives you the remainer: (x + y * width) % width = x.
  4. TheUnbeliever

    Alternatives to RANSAC

    Can you clarify your system some more? RANSAC is a general robust estimator. It's entirely possible to use it to extract multiple models from a dataset by e.g. running it repeatedly on the 'leftovers'. EDIT: Didn't realize this thread was from March.
  5. TheUnbeliever

    NettleBox2 Holographic surface.

    I ruled this out because of the limited viewing angles if the glasses were polarized, but obviously this wouldn't be a problem with active shutter, which I didn't think of for some reason...
  6. TheUnbeliever

    NettleBox2 Holographic surface.

    Smooth. Do you have a paper? How do you plan on handling perspective for multiple players?
  7. I'm finding it difficult to follow what you're saying, but it sounds like it could be greatly simplified by just introducing iterator invalidation akin to that in the standard library. If I [font=courier new,courier,monospace]erase[/font] from a container, my existing iterator is invalidated - I must instead use the one returned by the function.
  8. TheUnbeliever

    Understanding the D* lite whitepaper

    Three affirmatives. :-) (I hadn't even noticed the dot first time!)
  9. TheUnbeliever

    Understanding the D* lite whitepaper

    Not quite. That would be written 'argmin', whereas we just have 'min'. This is the minimum value of g(s') + c(s',u) for any predecessor s', not the predecessor itself. When comparing: U.TopKey < calculateKeys, ie: (X1, Y1) < (X2, Y2), is this (X1 < X2) AND (Y1 < Y2), or (X1 < X2) OR (Y1 < Y2), or something else entirely?[/quote] Explained from around the middle of the first paragraph on page 478, starting at 'keys are compared according to a lexicographic ordering.'
  10. TheUnbeliever

    Understanding the D* lite whitepaper

    It returns a pair or 2D vector; the authors have opted for [a;b] over (a,b) to improve legibility. The other line I had trouble with was this one: min [sub]s' (is element of) pred(s)[/sub] (g(s'), c(s', s)). I don't get that one at all. The s' in pred() with the smallest g or c?[/quote] I don't see this line. I do see min [sub]s' (is element of) pred(s)[/sub] (g(s') + c(s', s)) - i.e. the minimum of the expression on the right over all predecessors of s. EDIT: slightly clearer wording.
  11. TheUnbeliever

    simple c++ bool question

    It's still polling - this is actually (a minor variation on) the solution proposed several times earlier in the thread.
  12. TheUnbeliever

    Is this the future of IDEs?

    Relevant: Light Table (the Kickstarter's finished now - it got funded, and is in this summer's Y Combinator; there's more details available elsewhere on the site)
  13. TheUnbeliever

    Another Bezier path question

    If you do as I suggested and make sure your curves are continuous, you should get what you want. Assuming I've done things right, I get C[sub]2 [/sub]continuity between curves p and q (i.e. position, velocity and acceleration continuous across the join) requiring: q[sub]0[/sub] = p[sub]3[/sub] q[sub]1[/sub] = 2p[sub]3[/sub] - p[sub]2[/sub] q[sub]2[/sub] = p[sub]1[/sub] - 4p[sub]2[/sub] + 4p[sub]3[/sub] Working: [spoiler]p(t) = (1-t)[sup]3[/sup]p0 + 3t(1-t)[sup]2[/sup]p[sub]1[/sub] + 3t[sup]2[/sup](1-t)p[sub]2[/sub] + t[sup]3[/sup]p[sub]3[/sub] p'(t) = -3[(t-1)[sup]2[/sup]p[sub]0[/sub] - (3t[sup]2[/sup]-4t+1)p[sub]1[/sub] + t(3t-2)p[sub]2[/sub] - t[sup]2[/sup]p[sub]3[/sub]] p''(t) = -6[(t-1)p0 - (3t-2)p[sub]1[/sub] + (3t-1)p[sub]2[/sub] - tp[sub]3[/sub]] C[sub]0[/sub]: p[sub]3[/sub] = q[sub]0[/sub] C[sub]1[/sub]: p[sub]2[/sub] - p[sub]3[/sub] = q[sub]0[/sub] - q[sub]1[/sub] p[sub]2[/sub] - 2p[sub]3[/sub] = -q[sub]1[/sub] q[sub]1[/sub] = 2p[sub]3[/sub] - p[sub]2[/sub] C[sub]2[/sub]: -p[sub]1[/sub] + 2p[sub]2[/sub] - p[sub]3[/sub] = -q[sub]0[/sub] + 2q[sub]1[/sub] - q[sub]2[/sub] = -p[sub]3[/sub] + 4p[sub]3[/sub] - 2p[sub]2[/sub] - q[sub]2[/sub] -p[sub]1[/sub] + 4p[sub]2[/sub] - 4p[sub]3[/sub] = -q[sub]2[/sub] q[sub]2[/sub] = p[sub]1[/sub] - 4p[sub]2[/sub] + 4p[sub]3[/sub] [/spoiler] How do you determine the curve length? If you scale your steps in t by the arc length of the current curve, that should work.
  14. TheUnbeliever

    Another Bezier path question

    If the curves are continuous (which you can arrange by equating however many derivatives as is appropriate for the level of 'smoothness' you want and solving the resulting system of equations) and you carry whatever is left of any step into the next curve, the joins should be imperceptible.
  15. TheUnbeliever

    3D Modelling using spreadsheets

    I think, if you laid your spreadsheet out appropriately, you could save as CSV and the only conversion that would be involved would be a fairly simple regex.
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