1

我的代码：

#include <stdio.h>
main()
{
    const int x = 10;
    int *p;
    p=&x;
    *p=20;
    printf("Value at p: %d \n",*p);
    printf("Value at x: %d", x);
}

我得到的输出是：

p 值：20
x 值：20

因此，常量变量的值被改变。这是使用指针的缺点之一吗？

4

5 回答 5

2

You used a int* to point to a const int. you should get:

 error: invalid conversion from ‘const int*’ to ‘int*’

when you do:

p = &x;

You probably needs to update your compiler, a decent compiler should have told you this error or at least gave you warning about this.

于 2013-05-06T01:33:45.833 回答
0
于 2013-05-06T01:33:32.033 回答
0

That's because you are using the C language in the wrong way and the compiler let you compile this code while giving only a warning

warning: assignment discards ‘const’ qualifier from pointer target type [enabled by default]

And there are answers for that on SO, such as warning: assignment discards qualifiers from pointer target type

于 2013-05-06T01:34:50.150 回答
0

Any decent compiler will tell you that you are discarding the const qualifier.

C assumes that the programmer is always right, hence it's your choice to ignore your compiler's warnings or not. As usual, this is not a disadvantage as long as you know what you're doing!

于 2013-05-06T01:35:19.913 回答
0

正如其他答案所指出的那样，编写一个程序来尝试修改这样的const限定变量会导致程序具有未定义的行为。这意味着您的程序可以做任何事情——例如，当我在启用优化的情况下编译您的程序时，我会看到以下输出：

Value at p: 20
Value at x: 10

..如果我将static限定符添加到变量中，x那么程序会在运行时崩溃。

于 2013-05-06T01:51:30.380 回答