超出 C++ 的界限和未定义的行为 [英] Out of the bounds in C++ and undefined behaviour

问题描述

我知道在 C++ 中访问缓冲区边界是未定义的行为.
这是来自 cppreference 的示例:

I know that in c++ access out of buffer bounds is undefined behaviour.
Here is example from cppreference:

int table[4] = {};
bool exists_in_table(int v)
{
    // return true in one of the first 4 iterations or UB due to out-of-bounds access
    for (int i = 0; i <= 4; i++) {
        if (table[i] == v) return true;
    }
    return false;
}

但是，我在 C++ 标准中找不到相应的段落.
任何人都可以指出我在标准中解释这种情况的具体段落吗?

But, I can't find according paragraph in c++ standard.
Can anyone point me out on concrete paragraph in standard where such case is explained?

推荐答案

这是未定义的行为.我们可以并列几段经文来证实这一点.首先，我不会明确证明，table[4] 是 *(table + 4).我们只需要问自己指针值table + 4 的属性以及它与间接运算符的要求有何关系.

It's undefined behavior. We can juxtapose a couple of passages to be convinced of it. First, and I won't explicitly prove it, table[4] is *(table + 4). We need only ask ourselves the properties of the pointer value table + 4 and how it relates to the requirements of the indirection operator.

在指针上，我们有这样一段话:

On the pointer, we have this passage:

[basic.compound]

3 指针类型的每个值是以下之一:

3 Every value of pointer type is one of the following:

• 指向对象或函数的指针(据说该指针指向对象或函数)，或
• 超过对象末尾的指针([expr.add])，或
• 该类型的空指针值，或
• 无效的指针值.

我们的指针是第二个项目符号的类型，而不是第一个.至于间接运算符:

Our pointer is of the second bullet's type, not the first. As for the indirection operator:

[expr.unary.op]

1 一元 * 运算符执行间接寻址:应用它的表达式应该是指向对象类型的指针，或指向函数类型的指针，结果是引用表达式指向的对象或函数的左值.如果表达式的类型为pointer to T"，则结果的类型为T".

1 The unary * operator performs indirection: the expression to which it is applied shall be a pointer to an object type, or a pointer to a function type and the result is an lvalue referring to the object or function to which the expression points. If the type of the expression is "pointer to T", the type of the result is "T".

我希望通过阅读本段可以明显看出，该操作是为上一段中第一个项目符号描述的类别的指针定义的.

I hope it's obvious from reading this paragraph that the operation is defined for a pointer of the category described by the first bullet in the preceding paragraph.

因此，我们将操作应用于未定义其行为的指针值.结果是未定义的行为.

So we apply an operation to a pointer value for which its behavior is not defined. The result is undefined behavior.

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