在 C++ 中修改 const int

running the following code shows that &x=ptr, so how come x and *ptr are not equal?

const int x=10;
int* ptr =(int*) &x;
*ptr = (*ptr)+1;

cout << &x << " " << x << "  " << ptr <<"  " <<*ptr;  //output : 0012FF60 10  0012FF60  11

解决方案

The C++ implementation is only required to make a program work if you obey the rules. You violated the rules. The C++ implementation likely behaved this way:

• Because x is declared const, the C++ implementation knows its value cannot change as long as you obey the rules. So, wherever x is used, the C++ implementation uses 10 without bothering to check whether x has changed.
• Because *ptr points to a non-const int, stores to it and reads from it are actually performed. These "work" because the memory it points to (where x is represented) is not actually marked read-only by the operating system. Thus, you are able to make modifications in spite of the fact that you are not supposed to.

Observe that the behavior of the C++ implementation would work if you obeyed the rules. If you had not modified x, then using 10 for x wherever it appeared would have worked normally. Or, if you had not declared x to be const, then the C++ implementation would not have assumed it would always be 10, so it would get the changed value whenever x was accessed. This is all the C++ standard requires of an implementation: That it work if you follow the rules.

When you do not follow the rules, a C++ implementation may break in seemingly inconsistent ways.