C++ 将 std::list<>::iterator 的值获取到指针？
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Getting value of std::list<>::iterator to pointer?
提问by Onedayitwillmake
How can i loop thru a stl::List and store the value of one of the objects for use later in the function?
我如何通过 stl::List 循环并存储其中一个对象的值以供稍后在函数中使用？
Particle *closestParticle;
for(list<Particle>::iterator p1 = mParticles.begin(); p1 != mParticles.end(); ++p1 )
{
// Extra stuff removed
closestParticle = p1; // fails to compile (edit from comments)
}
回答by sbi
Either
任何一个
Particle *closestParticle;
for(list<Particle>::iterator it=mParticles.begin(); it!=mParticles.end(); ++it)
{
// Extra stuff removed
closestParticle = &*it;
}
or
或者
list<Particle>::iterator closestParticle;
for(list<Particle>::iterator it=mParticles.begin(); it!=mParticles.end(); ++it )
{
// Extra stuff removed
closestParticle = it;
}
or
或者
inline list<Particle>::iterator findClosestParticle(list<Particle>& pl)
{
for(list<Particle>::iterator it=pl.begin(); it!=pl.end(); ++it )
{
// Extra stuff removed
return it;
}
return pl.end();
}
or
或者
template< typename It >
inline It findClosestParticle(It begin, It end)
{
while(begin != end )
{
// Extra stuff removed
return begin;
++begin;
}
return end;
}
These are sorted in increasing personal preference. :)
这些按个人喜好增加排序。 :)
回答by rlbond
For a list, the only way to invalidate an iterator is to eraseit. So I suspect you're calling list.erase(p1)at some point in the loop. You need to make a copy of the iterator, move p1back one, and then erase the copy.
对于 a list，使迭代器无效的唯一方法就是使用erase它。所以我怀疑你list.erase(p1)在循环中的某个时刻打电话。您需要制作迭代器的副本，移p1回一个，然后删除副本。
EDIT: Oh wait, did you mean it doesn't compile? If so, see @sbi's answer. But you really need to word your question in a good way. What is your compile error? Or does it fail at run-time? In this case, however, I believe you mean a compile error.
编辑：哦等等，你的意思是它不能编译吗？如果是这样，请参阅@sbi 的回答。但是你真的需要以一种好的方式来表达你的问题。你的编译错误是什么？或者它在运行时失败？但是，在这种情况下，我相信您的意思是编译错误。
回答by Carl
I'm not an expert on the STL, but I believe the reason it fails to compile is because an iterator is an object that points to another object. In other words, an iterator is a generalization of a pointer. So to do what you'd want with minimal changes to your code, you would first need to de-reference the iterator to get at the value it contains. You'd then use the '&' to get its address and would then assign that address to your pointer variable. This is why ptr=&*it; works.
我不是 STL 的专家，但我相信它无法编译的原因是因为迭代器是一个指向另一个对象的对象。换句话说，迭代器是指针的泛化。因此，要在对代码进行最少更改的情况下执行您想要的操作，您首先需要取消引用迭代器以获取它包含的值。然后您将使用 '&' 获取其地址，然后将该地址分配给您的指针变量。这就是为什么 ptr=&*it; 作品。
