C语言const研究

刚刚写了一段这样子的代码：

#include <stdio.h>

int main()
{
    const int two = 2;
    int *one = &two;
    *one =1;

    printf("%d\n", two);

    return 0;
}

在ANSI C中运行通过了。 结果发现竟然成功输出了=。=、 看来C语言的常量限制仅陷于某个变量名啊。

1

在K & R C中则不能通过，因为不相容问题。

C++也是不行的:

#include <iostream>
using namespace std;

int main()
{
    const int two = 2;
    int *one = const_cast<int *>(&two);
    *one = 1;

    printf("%p %p\n", one, &two);
    printf("%d\n", two);
    printf("%d\n", *one);

    return 0;
}

运行结果：

0x7fff65150a54 0x7fff65150a54
2
1

诡异地出现同样地址不一样的值!

进一步修改代码：

#include <iostream>
using namespace std;

int main()
{
    const int two = 2;
    int *one = const_cast<int *>(&two);

    printf("%d\n", *one);
    *one = 1;

    printf("%p %p\n", one, &two);
    printf("%d\n", two);
    printf("%d\n", *one);

    int *three = const_cast<int *>(&two);
    printf("%d\n", *three);

    return 0;
}

结果：

2
0x7fff669b6a54 0x7fff669b6a54
2
1
1

有一种常量和变量完全剥离的感觉！ 于是在stackoverflow上发现下面这段子：

In C++ language it is illegal to attempt to modify constant objects. Such an attempt results in undefined behavior.

In your program the *p = 2 assignment attempts to modify a constant object a. The behavior is undefined. The weird output you observe is exactly that: undefined behavior.

There's no meaningful explanation for undefined behavior.

(Your compiler probably translated the cout << a; statement into cout << 1;, since the value of a cannot legally change. So, regardless of what you do to your a, 1 will always be printed.)