我们可以修改 const 变量的值吗？

Question

来自 this article .

Another use for declaring a variable as register and const is to inhibit any non-local change of that variable, even trough taking its address and then casting the pointer. Even if you think that you yourself would never do this, once you pass a pointer (even with a const attribute) to some other function, you can never be sure that this might be malicious and change the variable under your feet.

我不明白我们如何通过指针修改const 变量的值。这不是未定义的行为吗？

const int a = 81;
int *p = (int *)&a;
*p = 42; /* not allowed */

Answer 1

作者的观点是，使用 register 存储类声明变量会阻止您获取其地址，因此无法将其传递给可能通过丢弃 const 来更改其值的函数。

void bad_func(const int *p) {
    int *q = (int *) p;            // casting away const
    *q = 42;                       // potential undefined behaviour
}

void my_func() {
    int i = 4;
    const int j = 5;
    register const int k = 6;
    bad_func(&i);                  // ugly but allowed
    bad_func(&j);                  // oops - undefined behaviour invoked
    bad_func(&k);                  // constraint violation; diagnostic required
}

通过将潜在的 UB 更改为约束违规，需要进行诊断并且(需要)在编译时诊断错误::

c11

5.1.1.3 Diagnostics

1 - A conforming implementation shall produce at least one diagnostic message [...] if a preprocessing translation unit or translation unit contains a violation of any syntax rule or constraint, even if the behavior is also explicitly specified as undefined or implementation-defined.

6.5.3.2 Address and indirection operators

Constraints

1 - The operand of the unary & operator shall be [...] an lvalue that designates an object that [...] is not declared with the register storage-class specifier.

请注意，register 数组对象上的数组到指针衰减是未定义的行为，不需要进行诊断 (6.3.2.1:3)。

另请注意，在 C++ 中，register 左值 允许获取地址，其中 register 只是一个优化器提示(并且已弃用一个)。