关于使用指针修改 const 变量的困惑

Question

以下示例使我的理解更加困惑。我无法理解如何在本地修改 const 变量。请帮助我理解相同的内容。

 /* Compile code without optimization option */
 // volatile.c
 #include <stdio.h>
 int main(void)
 {
     const int local = 10;
     int *ptr = (int*) &local;

     printf("Initial value of local : %d \n", local);

     *ptr = 100;

     printf("Modified value of local: %d \n", local);

     return 0;
}

$ gcc volatile.c -o volatile –save-temps

$./ volatile

本地初始值:10

local的修改值:100

Answer 1

这就是undefined behavior如果我们查看 C99 草案标准部分 6.7.3 Type qualifiers paragraph 4 它说:

If an attempt is made to modify an object defined with a const-qualified type through use of an lvalue with non-const-qualified type, the behavior is undefined. If an attempt is made to refer to an object defined with a volatile-qualified type through use of an lvalue with non-volatile-qualified type, the behavior is undefined.¹¹⁵⁾

所以你不能对结果有任何期望，你不应该这样做。

如果我们看第 2 段，它说:

The properties associated with qualified types are meaningful only for expressions that are lvalues.¹¹⁴⁾

和脚注 114 说:

The implementation may place a const object that is not volatile in a read-only region of storage. Moreover, the implementation need not allocate storage for such an object if its address is never used.

一般来说，实现不必使常量变量只读，但它可能但正如 R.. 指出的那样，将自动变量放入只读存储器中会很难。