在STL迭代器的上下文中，单数和非奇异值是什么？ [英] What is singular and non-singular values in the context of STL iterators?

问题描述

C ++标准（2003）中的§24.1/ 5一节中的

只是作为数组的常规指针
保证有一个指针
值指向超过数组的最后一个元素
，所以对于任何迭代器类型
，有一个迭代器值指向
超过最后一个元素的一个
对应的容器。这些值
称为过期值。定义了
表达式* i的迭代器i的值
称为
dereferenceable。库从不
假定过去的结束值是
dereferenceable。 迭代器也可以
有奇异值，而不是
与任何容器相关联。
[示例：声明
后未初始化的指针x与int *
x;），x必须总是假定有
a指针的奇异值。]
大多数表达式的结果是
未定义奇异值;
唯一的例外是一个
非奇异值赋值给一个迭代器，
拥有一个奇异值。在这种情况下
奇异值被覆盖的
与任何其他值相同的方式。
可引用的值总是
nonsingular。

我真的不明白粗体？

• 什么是奇异值和非奇异值？如何定义它们？


迭代器也可以有不与任何容器相关联的奇异值。。

p>

我想这是它的定义。

因为如果它们不会，那么解除引用它们将是未定义的行为。

The section §24.1/5 from the C++ Standard (2003) reads,

Just as a regular pointer to an array guarantees that there is a pointer value pointing past the last element of the array, so for any iterator type there is an iterator value that points past the last element of a corresponding container. These values are called past-the-end values. Values of an iterator i for which the expression *i is defined are called dereferenceable. The library never assumes that past-the-end values are dereferenceable. Iterators can also have singular values that are not associated with any container. [Example: After the declaration of an uninitialized pointer x (as with int* x;), x must always be assumed to have a singular value of a pointer.] Results of most expressions are undefined for singular values; the only exception is an assignment of a non-singular value to an iterator that holds a singular value. In this case the singular value is overwritten the same way as any other value. Dereferenceable values are always nonsingular.

I couldn't really understand the text shown in bold?

• What is singular value and nonsingular value? How are they defined? And where?
• How and why dereferenceable values are always nonsingular?

解决方案

Iterators can also have singular values that are not associated with any container.

I suppose that's its definition.

How and why dereferenceable values are always nonsingular?

Because if they wouldn't, dereferencing them would be undefined behavior.

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