2 回答
This is totally flawed. You create and destroy a temporary string every time you call Test(). Any attempt to access memory using pointer returned by Test().c_str() after temporary was destroyed makes no sense - memory was freed already. It MIGHT have the old values (if nothing is written there before the access), but it might have anything as well (if it is reused before the access). It's Undefined Behavior.
In case of VC++ it is overwritten once and is not in other cases. With GCC - it's never overwritten. But this is pure chance. Once again - it's UB.
You have undefined behavior. The std::string returned by Test() is a temporary and the pointer returned by c_str() (stored in j) is no longer valid after the lifetime of the temporary ends. This means that anything can happen. The array the pointer points to may contain garbage, it may be the original string or the implementation may have null terminated the beginning of it. Accessing it may cause a segmentation fault or it may allow you to access the original string data. This can and usually does vary between different compilers and implementations of the standard library.
char *j = const_cast<char*>(Test().c_str());
// The contents pointed to by j are no longer valid and access that content
// is undefined behavior
cout << "address:"<< (unsigned)Test().c_str() << endl;
The address is different between calls to Test() because it returns a temporary each time you call it. Some compilers may optimize this and/or the allocation of data may get the same block of memory but it is not guaranteed to be the same.