C++ free all memory used by struct

问题:

Quick question; I've googled around and found some answers already, but I'm a bit paranoid so I want to be sure.

Consider this situation:

struct CoordLocation
{
    float X;
    float Y;
    float Z;
};

int main()
{
    CoordLocation *coord = new CoordLocation();
    delete coord;
    return 0;
}

Will calling delete also clear the memory used by the fields X, Y, Z? Some answers I found mentioned that I'd just delete the POINTER, not the actually referenced object this way. What if...

struct CoordLocation
{
    float *X;
    float *Y;
    float *Z;
};

int main()
{
    CoordLocation *coord = new CoordLocation();
    delete coord;
    return 0;
}

And what if I manually free the memory for each object inside the struct's constructor/destructor?

struct CoordLocation
{
    CoordLocation()
    {
         *X = new float;
         *Y = new float;
         *Z = new float;
    }
    ~CoordLocation()
    {
         delete X; delete Y; delete Z;
    }
    float *X;
    float *Y;
    float *Z;
};

int main()
{
    CoordLocation *coord = new CoordLocation();
    delete coord;
    return 0;
}

I noticed that for a simple situation such as:

   float *a = new float;
   *a = 5.0f;
   printf("%f", *a);
   delete a;
   printf("%f", &a);

printf would print 5.0, so the variable pointed to by a is not exactly destroyed.

So my question is: How can I reliably free (as in no memory leaks) ALL the memory used by the struct in this case?

struct CoordLocation
{
    float X;
    float Y;
    float Z;
};

int main()
{
    CoordLocation *coord = new CoordLocation();
    delete coord;
    return 0;
}

Thanks!

回答1:

You only need to delete memory you allocate with new.

printf would print 5.0, so the variable pointed to by a is not exactly destroyed.

You're actually running into undefined behavior. Although the value is still there, the memory was released and can be reused.

So the following:

struct CoordLocation
{
    float X;
    float Y;
    float Z;
};

can't create a memory leak if you omit the destructor.

Your next snippet:

struct CoordLocation
{
    float *X;
    float *Y;
    float *Z;
};

int main()
{
    CoordLocation *coord = new CoordLocation();
    delete coord;
    return 0;
}

can potentially create a memory leak, but not as it is. The following will:

int main()
{
    CoordLocation *coord = new CoordLocation();
    coord->X = new float();
    delete coord;
    return 0;
}

Your third example

struct CoordLocation
{
    CoordLocation()
    {
         *X = new float;
         *Y = new float;
         *Z = new float;
    }
    ~CoordLocation()
    {
         delete X; delete Y; delete Z;
    }
    float *X;
    float *Y;
    float *Z;
};

int main()
{
    CoordLocation *coord = new CoordLocation();
    delete coord;
    return 0;
}

won't create a memory leak because you free all the memory that you allocate. If you were to omit the destructor or forget to call delete coord;, they you'd have a memory leak.

A good rule of thumb: call a delete for every new and a delete[] for every new[] and you're safe.

回答2:

In this example:

struct CoordLocation
{
    float X;
    float Y;
    float Z;
};

int main()
{
    CoordLocation *coord = new CoordLocation();
    delete coord;
    return 0;
}

The memory is freed. There are no leaks in that code. On the example you gave, where your struct contained pointers, as long as you free them in the destructor(as you did in your example), there will be no leaks.

In this snippet:

float *a = new float;
*a = 5.0f;
printf("%f", *a);
delete a;
printf("%f", &a);

The value of a stays the same after the delete, since delete(ing) a pointer will only mark a memory address as "freed", it will not modify its contents. Accessing freed pointers is undefined behaviour.

回答3:

The pointer itself is allocated with automatic storage duration, there is nothing to free there. The struct is those three fields, which are freed when you call delete. You only call delete on something that was returned by new.

When you allocate something you allocate enough memory to hold it, which means enough memory to hold all of its fields (and some housekeeping memory that is implementation specific, but you don't worry about that). When you delete it you free the same amount of memory that you allocated.

One thing to note however is a case like this (in C++ you wouldn't create a type like this, but the example is relevant):

struct Foo {
    char *str;
};

Foo *f = new Foo();
f->str = new char[10];

delete f;

In this case you have a leak. You delete f, which consists of enough memory to hold a single char*, but what that char* points to has been dynamically allocated as well. So, you need to free it too:

delete f->str;
delete f;

Again, in C++ you would probably not design a type this way, instead favoring types like std::string and principles such as RAII, but the example is relevant.