是下面的代码使用std :: set" legal&quot ;? [英] Is the following code using std::set "legal"?

问题描述

我有这个代码：

set<int>::iterator new_end = 
                   set_difference(set1.begin(), set1.end(),
                                  set2.begin(), set2.end(),
                                  set1.begin());
set1.erase(new_end, set1.end);

它在visual studio中编译并运行良好。但是，在上一个问题中，人们表示 set 的迭代器应该是 const 。我在标准中没有看到这样的东西。有人可以告诉我它在哪里说，或者如果这是明确定义的行为？

It compiles and runs fine in visual studio. However, in a previous question, people stated that a set's iterators are supposed to be const. I don't see anything like that in the standard. Can someone tell me where it says that, or if this is well-defined behavior?

如果不是，请提供代码，我需要做的。有没有办法做到这一点，而不创建一个临时集？

If it's not, please provide code that does what I need. Is there a way to do this without creating a temporary set?

推荐答案

您的代码违反了一些不变量 set_difference 。从 Josuttis Book 的第420页起：

Your code violates a couple of the invariants for set_difference. From page 420 of the Josuttis Book:

• 调用者必须确保目标范围足够大或使用插入迭代器。


• 目标范围不应与源范围重叠。

您尝试回写第一个集合，这是不允许的。您需要编写除源范围以外的其他位置 - 我们可以使用第三组：

You're trying to write back over the first set, which is not allowed. You need to write someplace other than the source ranges - for that we can use a third set:

std::set<int> set3;
std::set_difference(set1.begin(), set1.end(),
                    set2.begin(), set2.end(),
                    std::inserter(set3, set3.begin()));

std :: insertionter的第二个参数是元素应插入的位置的提示。这只是一个提示，但是，放心，元素将最终在正确的地方。 set3 最初为空，因此 begin（）是我们可以提供的唯一提示。

The second argument to std::inserter is a hint for where the elements should be inserted. It's only a hint, however, rest assured that the elements will end up in the right place. set3 is initially empty, so begin() is about the only hint we can give.

调用 set_difference 后， set3 将包含您尝试创建的 set1 包含在原始代码中。您可以使用 set1 继续使用 set3 或 swap 如果您愿意。

After the call to set_difference, set3 will contain what you tried to make set1 contain in your original code. You can go on using set3 or swap it with set1 if you prefer.

更新：

如果你只想删除 set2 中出现的 set1 中的所有元素，你可以尝试：

I'm not sure about the performance of this, but if you just want to remove all elements from set1 that appear in set2, you can try:

for (std::set<int>::iterator i = set2.begin(); i != set2.end(); ++i)
{
    set1.erase(*i);
}

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