在C ++中修改const int [英] Modifying a const int in C++

问题描述

运行以下代码显示& x = ptr，那么x和* ptr的值如何不相等？

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

只有在遵守规则的情况下才能使程序工作。您违反了规则。 C ++实现可能表现如下：

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 $ c> const ，C ++实现知道它的值不能改变，只要你遵守规则。因此，在使用 x 的任何地方，C ++实现使用10而无需检查 x 是否已更改。 >
  
• 因为 * ptr 指向非常量 int ，存储并读取从它实际上执行。这些工作因为它指向的内存（其中表示 x ）并不实际被操作系统标记为只读。

• 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，C ++的行为是非常重要的，如果您遵守规则，则实施工作。如果你没有修改 x ，那么对于 x 使用10就可以正常工作。或者，如果你没有声明 x 是 const ，那么C ++实现不会假设它总是10，所以当访问 x 时，它将获得更改的值。这是所有的C ++标准要求的一个实现：如果你遵守规则，它工作。

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.

当你不遵守规则，C ++实现可能会打破看似不一致的方式。

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

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