问题描述
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
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 declaredconst
, the C++ implementation knows its value cannot change as long as you obey the rules. So, whereverx
is used, the C++ implementation uses 10 without bothering to check whetherx
has changed. - Because
*ptr
points to a non-constint
, stores to it and reads from it are actually performed. These "work" because the memory it points to (wherex
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.
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