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Home/ Questions/Q 7766219
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Editorial Team
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Editorial Team
Asked: June 1, 20262026-06-01T15:21:33+00:00 2026-06-01T15:21:33+00:00

I have a relatively simple algorithm that walks an std::vector looking for two neighbouring

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I have a relatively simple algorithm that walks an std::vector looking for two neighbouring tuples. Once the tuples left and right of the X value are found I can interpolate between them. Somehow this works:

  std::vector<LutTuple*>::iterator tuple_it;
  LutTuple* left = NULL;
  LutTuple* right = NULL;
  bool found = 0;

  // Only iterate as long as the points are not found
  for(tuple_it = lut.begin(); (tuple_it != lut.end() && !found); tuple_it++) {
    // If the tuple is less than r2 we found the first element
    if((*tuple_it)->r < r) {
        left = *tuple_it;
    }
    if ((*tuple_it)->r > r) {
        right = *tuple_it;
    }
    if(left && right) {
        found = 1;
    }
  }

while this:

  std::vector<LutTuple*>::iterator tuple_it;
  LutTuple* left = NULL;
  LutTuple* right = NULL;

  // Only iterate as long as the points are not found
  for(tuple_it = lut.begin(); tuple_it != lut.end() && !left && !right; tuple_it++) {
    // If the tuple is less than r2 we found the first element
    if((*tuple_it)->r < r) {
        left = *tuple_it;
    }
    if ((*tuple_it)->r > r) {
        right = *tuple_it;
    }
  }

does not. Why is that? I’d expect two NULL ptrs like this to evaluate to true together when negated.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-01T15:21:34+00:00Added an answer on June 1, 2026 at 3:21 pm

    There is a logical issue.

    In the first snippet you have (essentially) !(left && right).

    In the second snippet you have: !left && !right.

    Those are not equivalent.

    If you build the truth table, you will realize that !(left && right) is equivalent to (!left || !right).

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