c - 被误解的代码示例

Question

我正在读这本书http://publications.gbdirect.co.uk/c_book/chapter8/const_and_volatile.html我停在其中一个例子上。我认为这是不正确的。我认为，没有未定义的行为。我错了吗？在这里:

Taking the address of a data object of a type which isn't const and putting it into a pointer to the const-qualified version of the same type is both safe and explicitly permitted; you will be able to use the pointer to inspect the object, but not modify it. Putting the address of a const type into a pointer to the unqualified type is much more dangerous and consequently prohibited (although you can get around this by using a cast). Here is an example:

#include <stdio.h>
#include <stdlib.h>

main(){
        int i;
        const int ci = 123;

        /* declare a pointer to a const.. */
        const int *cpi;

        /* ordinary pointer to a non-const */
        int *ncpi;

        cpi = &ci;
        ncpi = &i;

        /*
         * this is allowed
         */
        cpi = ncpi;

        /*
         * this needs a cast
         * because it is usually a big mistake,
         * see what it permits below.
         */
        ncpi = (int *)cpi;

        /*
         * now to get undefined behaviour...
         * modify a const through a pointer
         */
        *ncpi = 0;

        exit(EXIT_SUCCESS);
}

Example 8.3 As the example shows, it is possible to take the address of a constant object, > generate a pointer to a non-constant, then use the new pointer. This is an error in your program and results in undefined behaviour.

在这个例子中，ncpi 最终指向i，而不是ci。所以我认为这使得这个例子不正确——在通过指针修改非 const 变量时没有未定义的行为。你同意吗？

Answer 1

我同意:这是一个有缺陷的例子。代码本身表现出定义的行为。

最终赋值前的注释，*ncpi = 0;，与代码不一致。可能作者打算做一些不同的事情。

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我的第一 react 是好像代码覆盖了一个const: 我修改了我的答案。