标准引用:对`vector :: operator []`的超出范围调用是否会调用未定义行为? [英] Standard citation: Does an out of range call to `vector::operator[]` invoke Undefined Behavior?

问题描述

考虑:

#include <vector>

int main()
{
  std::vector <int> v;
  v.reserve (100);
  v[1] = 42;
}

我知道上面的代码调用了Undefined Behavior，因为在v[1]的调用位置，根据v.size()的向量的大小为< 1.(为零.)

I am aware that the above code invokes Undefined Behavior, because at the call-site for v[1], the size of the vector according to v.size() is < 1. (It is zero.)

但是，我花了最后一个小时左右的时间浏览C ++标准，以直接或间接地澄清这实际上是未定义的行为.

However I have spent the last hour or so looking through the C++ Standard for a citation that either directly or indirectly clarifies that this is, in fact, Undefined Behavior.

也许我错过了显而易见的事情.您能否至少从C ++ 03以及可选的C ++ 11标准中提供引用?

Maybe I have missed the obvious. Can you provide a citation or citations from at least the C++03 and optionally the C++11 Standard which clears this up?

推荐答案

这是operator[]定义方式的产物.在§23.2.3[sequence.rqmts]表101中，operator[]的操作语义定义为*(a.begin() + n).因此，它是根据迭代器定义的.根据§24.2.1/5 [iterator.requirements.general]，将begin递增到end并对其取消引用是未定义的行为.

It's an outgrowth of how operator[] is defined. In §23.2.3 [sequence.rqmts] Table 101, the Operational Semantics of operator[] is defined as *(a.begin() + n). So it's defined in terms of iterators. And incrementing begin past end and dereferencing it is undefined behavior according to §24.2.1/5 [iterator.requirements.general].

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