修改 C++ 中的 const int [英] Modifying a const int in C++

问题描述

运行下面的代码显示&x=ptr，那么x和*ptr怎么不相等?

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

推荐答案

C++ 实现只需要您遵守规则才能使程序运行.你违反了规则.C++ 实现可能是这样的:

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:

• 因为 x 被声明为 const，C++ 实现知道只要你遵守规则，它的值就不会改变.因此，无论在何处使用 x，C++ 实现都会使用 10，而无需费心检查 x 是否已更改.
• 因为 *ptr 指向一个非常量 int，所以实际执行了存储和读取操作.这些工作"是因为它指向的内存(表示 x 的地方)实际上并未被操作系统标记为只读.因此，您可以进行修改，尽管您不应该这样做.

• 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.

请注意，如果您遵守规则，C++ 实现的行为将可以工作.如果您没有修改 x，那么在 x 出现的任何地方使用 10 都会正常工作.或者，如果您没有将 x 声明为 const，那么 C++ 实现不会假定它总是 10，因此每当 x 被访问.这就是 C++ 标准对实现的所有要求:如果你遵守规则，它就可以工作.

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.

当您不遵守规则时，C++ 实现可能会以看似不一致的方式中断.

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

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