是“向量迭代器+偏移量超出范围"吗?断言有用吗? [英] Is "vector iterator + offset out of range" assertion useful at all?

问题描述

这个完美的程序在 Visual Studio 2013 的调试模式下失败:

This perfectly good program fails in debug mode in Visual Studio 2013:

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

void main()
{
  vector<int> v = {3, 1, 4, 1, 5, 9, 2, 6, 5, 3};

  for (auto iFrom = v.cbegin(), iTo = iFrom+5; iFrom != v.cend(); iFrom = iTo, iTo += 5)
    cout << *max_element(iFrom, iTo) << '\n';
}

with vector iterator + offset out of range 断言失败.它失败是因为 iTo >v.cend()，这里是无害的.调试器测试迭代器的值有什么意义，它没有被取消引用?

with vector iterator + offset out of range assertion failure. It fails because iTo > v.cend(), which is harmless here. What is the point of debugger testing the value of iterator, which is not being dereferenced?

顺便说一句，我知道我可以将上面的循环重写为:

BTW, I know that I can rewrite the loop above as:

for (auto i = v.cbegin(); i != v.cend(); i += 5)
  cout << *max_element(i, i+5) << '\n';

但我试图用一段更复杂的现实生活代码制作一个简单的例子，其中计算新的迭代器值在计算上是昂贵的.

but I was trying to make a simple example out of a piece of more complex real-life code, where calculating the new iterator value is computationally expensive.

我也意识到可以更改 _ITERATOR_DEBUG_LEVEL 值来影响这种行为，但这会导致某些库的二进制版本出现问题，这些库是使用默认调试设置构建的.

I also realize that one can change _ITERATOR_DEBUG_LEVEL value to affect this behavior, but it creates problems with binary versions of some libraries, which are built with default debug settings.

推荐答案

这不是无害的...它是未定义的行为试图将迭代器移过 end()，您在第三次也是最后一次迭代中执行的操作.在使用 iTo += 5 执行此操作后，您立即以 iFrom != v.cend() 终止循环而不取消引用迭代器这一事实无关紧要.

It's not harmless... it's undefined behaviour to attempt to move the iterator past end(), which you do on your third and final iteration. The fact that immediately after doing so with iTo += 5 you terminate your loop as iFrom != v.cend() without dereferencing the iterator is not relevant.

如果确实需要效率，并且您准备好让银行押注元素数量是 5 的倍数:

If efficiency is really necessary and you're prepared the bet the bank on the number of elements being a multiple of 5:

for (auto iFrom = v.cbegin(), iTo = iFrom; iFrom != v.cend(); iFrom = iTo)
{
    iTo += 5;
    cout << *max_element(iFrom, iTo) << '\n';
}

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