删除它指向的动态分配的内存后使用指针时没有错误

Question

I'm learning about dynamic memory in C++. My question is why after removing the variable in the following code I don't get an error?

float* uf = new float(4.26);
delete uf;
cout << uf << '\n';
cout << *uf << '\n';

Answer 1

You don't get an error at compile time , because while the compiler tries to detect straightforward mistakes, it can't possibly know, at any given time, weather or not your pointer points to valid memory or not.

You may not get an error at run time , because even though the memory has been "freed", it is still memory. It's just not guaranteed to be yours anymore. This is the problem with undefined behavior: it is undefined; you never know exactly how a problem might manifest itself.

Answer 2

You mean because it compiles even though it's clearly wrong?

Arguably, in this case the compiler could produce a warning if it really wanted to, but then it's a simplified, unrealistic situation.

Consider the following example:

float* uf = new float(4.26);
delete uf;
if (random_condition_known_only_at_runtime()) {
    uf = new float(0.0);
}
cout << uf << '\n';
cout << *uf << '\n';

Or:

float* uf = new float(4.26);
if (user_input == 'x') {
    delete uf;
}
cout << uf << '\n';
cout << *uf << '\n';

Or consider concurrency; multiple threads may write to the same pointer.

The point is that real code will typically depend (directly or indirectly) a lot on I/O operations or other external state like this, making it impossible to know in advance, at compile time, whether the memory pointed to will have been deleted already.

---

Or do you mean because the program doesn't crash? That's because the C++ standard does not prescribe crashes. It instead refers to "undefined behaviour", which means that anything can happen, including random crashes or no effect at all. Trying to access memory which was already deleted is a classical example of such undefined behaviour.