Question

我有一个迭代器p和一个顶点curvePoints：

for (p = curvePoints.begin(); p != curvePoints.end(); p++) {
  p->x = (1 - u) * p->x + u * (p+1)->x;
  p->y = (1 - u) * p->y + u * (p+1)->y;
}

现在循环使用下一个索引对象的值;我怎样才能保证下一个值存在。换句话说，如何使循环条件类似(p+1) != curvePoints.end()或p != curvePoints.end() - 1。

Answer 1

如何将p != curvePoints.end()更改为(p != curvePoints.end()) && ((p + 1) != curvePoints.end())？

Answer 2

您可以在条件允许的情况下使用std::distance(p, curvePoints.end()) > 1。

http://www.cplusplus.com/reference/std/iterator/distance/

Answer 3

您可以缓存并重用p - s后继者的值：

auto p = curvePoints.begin();
if (p != curvePoints.end()) {
  auto pn = p; ++pn;
  for (; pn != curvePoints.end(); p = pn, ++pn) {
    ... // use pn instead of (p+1)
  }
}

注意：迭代器最好是预增量。

Answer 4

我这样做的方法是不与end()进行比较，除非序列为空。如果序列非空，您将与end() - 1进行比较，并可能缓存值：

for (auto p(points.begin()), 
          end(c.end() - (c.empty()? 0, 1));
     p != end; ++p) {
    ...
}

Answer 5

 for (p = curvePoints.begin(); p < curvePoints.end(); p++) {
          p->x = (1 - u) * p->x + u * (p+1)->x;
          p->y = (1 - u) * p->y + u * (p+1)->y;

在结束之前停止迭代器

5 个答案: