why vector pointer's memory automatically cleared - c++

BS: I've tested, the question below is VS2010 specific, in my test on Mingw, vector memory is leak, the
printf("%d\n", testV->size());
gives me '10', which shows it's a leak.
In codes like below:
class Test{
public:
std::vector<int> *num;
int *n;
};
Test *t = new Test();
t->num = new std::vector<int>;
t->num->assign(10,10);
t->n = new int[20];
t->n[0] = 12345;
int *testN = t->n;
std::vector<int> *testV;
testV = t->num;
delete t;
//I can print n's memory
printf("%d\n", testN[0]); //get 12345, so memory leaked here.
//I can't print num's memory; its size is 0. in release version, it's negative number.
printf("%d\n", testV->size());
After deleting t, I see that the memory num points to is cleared, in my VS10 environment, I could see its memory cleaned to 'ee fe ee fe....'.
But the memory of n stays the same.
In my opinion, all memory comes from keyword new, should be freed by delete, well, here the num (vector) is
not 'delete'd, but rather is cleared automatically.
I want to know how vector pointer is different from other ones?

Only t's memory is released, what the members num and n point to is leaked memory.
Your test is wrong, because you can't inspect t after you've deleted it, and there are no other ways to access those members (ergo the leak).

The debug runtime will overwrite memory it deletes with a marker like eefe to detect bugs like use-after-free. A release build will not do this.
You are probably just misreading the debug output. Both the vector* and the int* memory will still be around. The debugger is probably just trying to be smarter about vectors.

If you want to do this correctly, you need to delete [] t->n; and delete t->num; before you do delete t; - but a more correct way, using object orientation "correctly", would be to let t's constructor and destructor handle the memory allocation and freeing of n and num - with that, automatically, the memory is managed correctly for the class itself.
I would also point out that it's rarely right to use a pointer to a std::vector - the vector itself doesn't take up that much space, so you probably should use std::vector<int> num; instead of a pointer to it. This would also mean that num is automatically destroyed in the destructor without any additional code required.

Related

C++ Memory leaks vectors(?)

In C++, the importance of deallocating memory when the program is either exiting or no longer serves a purpose is important. So if this is allocation of a dynamic array
char** dynamicArr = nullptr;
for (int i = 0; i < x; i++) {
mapPtr[i] = new char[y];
}
and this is deallocation of a dynamic array
for (int i = 0; i < x; i++) {
delete[] mapPtr[i];
}
delete[] mapPtr;
However, when it comes to vectors, I noticed that my global vector with 0 elements inside seems to be causing some memory leaks.
I've read up on this link, a user commented that
No. The std::vector will automatically de-allocate the memory it uses
Screenshot of my debugging.
I have also tried these steps to clear the vector as well as make sure the vector inside the struct citySummInfo has shrunk to fit and clear hopefully not getting any memory leak but to no avail. Is there any way that I'm doing it wrong?
As what #JohnFilleau have mentioned
_CrtDumpMemoryLeaks() should be called at the point in the program where you want to see what is remaining on the heap. Since your
vectors are statically allocated, they will not have been destroyed at
the time you call this function.
_CrtDumpMemoryLeaks() is meant to place right before the program terminates, and since my vectors are statically allocated, it has not been deallocated at the time when _CrtDumpMemoryLeaks() has been called hence the "leaks".

Delete a vector without deleting the array

I'm trying to populate a vector of doubles in C++ and pass the associated array to Fortran. But I'm having trouble freeing the rest of the memory associated with the vector. I'd like to avoid copying. Here's what I have:
std::vector<double> *vec = new std::vector<double>();
(*vec).push_back(1.0);
(*vec).push_back(2.0);
*arr = (*vec).data(); //arr goes to Fortran
How do I delete vec while keeping arr intact? Is there a way to nullify the pointer to arr in vec so that I can then delete vec?
Update
I see that I didn't give enough information here. A couple things:
I'm actually calling a C++ function in Fortran using iso_c_binding
I don't know how large the vec needs to be. The vector class looks good for this situation
I might try Guillaume's suggestion eventually, but for now, I'm passing vec to the Fortran and calling another C++ function to delete it once I'm done with the data
You need to rethink your program design.
Somehow, somewhere, you need to keep an array alive while Fortran is using it. So whatever context you're using to access Fortran should probably be responsible for ownership of this array.
class fortran_context {
/*Blah blah blah whatever API you're using to access Fortran*/
void * arr;
std::vector<double> vec; //Don't allocate this as a pointer; makes no sense!
public:
fortran_context() {
arr = //Do whatever is necessary to setup Fortran stuff. I'm assuming your
//api has some kind of "get_array_pointer" function that you'll use.
}
~fortran_context() {
//Do the cleanup for the fortran stuff
}
//If you want to spend time figuring out a robust copy constructor, you may.
//Personally, I suspect it's better to just delete it, and make this object non-copyable.
fortran_context(fortran_context const&) = delete;
std::vector<double> & get_vector() {
return vec;
}
std::vector<double> const& get_vector() const {
return vec;
}
void assign_vector_to_array() {
*arr = vec.data();
}
void do_stuff_with_fortran() {
assign_vector_to_array();
//???
}
};
int main() {
fortran_context context;
auto & vec = context.get_vector();
vec.push_back(1.0);
vec.push_back(2.0);
context.do_stuff_with_fortran();
return 0;
} //Cleanup happens automatically due to correct implementation of ~fortran_context()
I've abstracted a lot of this because I don't know what API you're using to access Fortran, and I don't know what kind of work you're doing with this array. But this is, by far, the safest way to ensure that
The vector's allocated memory exists so long as you are doing stuff in Fortran
The memory associated with the vector will be cleaned up properly when you're done.
How do I delete vec while keeping arr intact? Is there a way to nullify the pointer to arr in vec so that I can then delete vec?
The library does not provide any built-in capability to do that. You have to do the bookkeeping work yourself.
Allocate memory for the data and copy data from the vector.
Send the data to FORTRAN.
Decide when it is safe to deallocate the data and then delete them.
// Fill up data in vec
std::vector<double> vec;
vec.push_back(1.0);
vec.push_back(2.0);
// Allocate memory for arr and copy the data from vec
double* arr = new double[vec.size()];
std::copy(vec.begin(), vec.end(), arr);
// use arr
// delete arr
delete [] arr;
What you are asking for is not possible. std::vector, being a well behaved template class will release the internals that it owns and manages when it is destroyed.
You will have to keep vector alive while you are using its contents, which makes perfect sense. Otherwise, you will have to make a copy.
Also, I don't see why you are allocating the vector on the heap, it doesn't seem needed at all.
How do I delete vec while keeping arr intact? Is there a way to nullify the pointer to arr in vec so that I can then delete vec?
You don't.
I think you misuse or misunderstood what vector is for. It is not meant to expose memory management of the underlying array, but to represent a dynamically sized array as a regular type.
If you need to explicitly manage memory, I'd suggest you to use std::unique_ptr<T[]>. Since unique pointers offers a way to manage memory and to release it's resource without deleting it, I think it's a good candidate to meet your needs.
auto arr = std::make_unique<double[]>(2);
arr[0] = 1.;
arr[1] = 2.;
auto data = arr.release();
// You have to manage `data` memory manually,
// since the unique pointer released it's resource.
// arr is null here
// data is a pointer to an array, must be deleted manually later.
delete[] data;

C++ Allocate Memory Without Activating Constructors

I'm reading in values from a file which I will store in memory as I read them in. I've read on here that the correct way to handle memory location in C++ is to always use new/delete, but if I do:
DataType* foo = new DataType[sizeof(DataType) * numDataTypes];
Then that's going to call the default constructor for each instance created, and I don't want that. I was going to do this:
DataType* foo;
char* tempBuffer=new char[sizeof(DataType) * numDataTypes];
foo=(DataType*) tempBuffer;
But I figured that would be something poo-poo'd for some kind of type-unsafeness. So what should I do?
And in researching for this question now I've seen that some people are saying arrays are bad and vectors are good. I was trying to use arrays more because I thought I was being a bad boy by filling my programs with (what I thought were) slower vectors. What should I be using???
Use vectors!!! Since you know the number of elements, make sure that you reserve the memory first (by calling myVector.reserve(numObjects) before you then insert the elements.).
By doing this, you will not call the default constructors of your class.
So use
std::vector<DataType> myVector; // does not reserve anything
...
myVector.reserve(numObjects); // tells vector to reserve memory
You can use ::operator new to allocate an arbitrarily sized hunk of memory.
DataType* foo = static_cast<DataType*>(::operator new(sizeof(DataType) * numDataTypes));
The main advantage of using ::operator new over malloc here is that it throws on failure and will integrate with any new_handlers etc. You'll need to clean up the memory with ::operator delete
::operator delete(foo);
Regular new Something will of course invoke the constructor, that's the point of new after all.
It is one thing to avoid extra constructions (e.g. default constructor) or to defer them for performance reasons, it is another to skip any constructor altogether. I get the impression you have code like
DataType dt;
read(fd, &dt, sizeof(dt));
If you're doing that, you're already throwing type safety out the window anyway.
Why are you trying to accomplish by not invoking the constructor?
You can allocate memory with new char[], call the constructor you want for each element in the array, and then everything will be type-safe. Read What are uses of the C++ construct "placement new"?
That's how std::vector works underneath, since it allocates a little extra memory for efficiency, but doesn't construct any objects in the extra memory until they're actually needed.
You should be using a vector. It will allow you to construct its contents one-by-one (via push_back or the like), which sounds like what you're wanting to do.
I think you shouldn't care about efficiency using vector if you will not insert new elements anywhere but at the end of the vector (since elements of vector are stored in a contiguous memory block).
vector<DataType> dataTypeVec(numDataTypes);
And as you've been told, your first line there contains a bug (no need to multiply by sizeof).
Building on what others have said, if you ran this program while piping in a text file of integers that would fill the data field of the below class, like:
./allocate < ints.txt
Then you can do:
#include <vector>
#include <iostream>
using namespace std;
class MyDataType {
public:
int dataField;
};
int main() {
const int TO_RESERVE = 10;
vector<MyDataType> everything;
everything.reserve( TO_RESERVE );
MyDataType temp;
while( cin >> temp.dataField ) {
everything.push_back( temp );
}
for( unsigned i = 0; i < everything.size(); i++ ) {
cout << everything[i].dataField;
if( i < everything.size() - 1 ) {
cout << ", ";
}
}
}
Which, for me with a list of 4 integers, gives:
5, 6, 2, 6

How to release the memory of structures pointed by an pointer-array?

I have a pointer array defined as some_struct * t_ptr[1000] which points to a structure some_struct.And some points of the point array are evaluated.e.g
some_struct * wpaper1 = new some_struct(); //memory leaks detected in this line
wpaper1->tAnswer = useransw;
wpaper1->tKey = key;
t_ptr[100] = wpaper1;
//there're wpaper2,wpaper3....
...
delete[] t_ptr; //release the pointers
The debug message says there're memory leaks detected in the first line code.So how can I release the memory of some_struct pointed by the t_ptr array?Do I have to use a loop to detect whether a element is evaluated and then delete it?And I'm using VS2008 on Windows.Thanks.
Your delete[] t_ptr would only be correct if you've allocated t_ptr on the heap, ala:
some_struct* t_ptr = new tpr[1000];
Then, the delete[] statement releases the memory for those 1000 pointers, but does nothing about any memory that the pointers themselves may refer to. To release that, you need to first loop over the t_ptr elements, deleting them one by one:
for (i = 0; i < 1000; ++i)
delete t_ptr[i];
delete[] t_ptr;
You must ensure the pointers are set to NULL initially, though deleting a NULL pointer is a no-operation so you don't need to check in the loop above.
It's a pain isn't it? That's why a very good guideline for new C++ programmers is to use vectors and smart pointers (e.g. from boost). Those types free themselves when they go out of scope or the program exits... you don't even have to think about it.
Yes, you have to iterate over t_ptr, and delete any elements that are non-NULL (assuming you've initialized the elements to NULL before you start allocating). You should be able to identify in your code where each delete is that matches each new.

What is the array form of 'delete'?

When I compiled a code using the array name as a pointer, and I deleted the array name using delete, I got a warning about deleting an array without using the array form (I don't remember the exact wording).
The basic code was:
int data[5];
delete data;
So, what's the array form of delete?
The array form of delete is:
delete [] data;
Edit: But as others have pointed out, you shouldn't be calling delete for data defined like this:
int data[5];
You should only call it when you allocate the memory using new like this:
int *data = new int[5];
You either want:
int *data = new int[5];
... // time passes, stuff happens to data[]
delete[] data;
or
int data[5];
... // time passes, stuff happens to data[]
// note no delete of data
The genera rule is: only apply delete to memory that came from new. If the array form of new was used, then you must use the array form of delete to match. If placement new was used, then you either never call delete at all, or use a matching placement delete.
Since the variable int data[5] is a statically allocated array, it cannot be passed to any form of the delete operator.
As the other have said, you must use the vector form of delete:
void some_func(size_t n)
{
int* data = new int[n];
. . . // do stuff with the array
delete [] data; // Explicitly free memory
}
Be very wary of this, because some compilers will not warn you.
Even better, there is very rarely any need for using vector new/delete. Consider whether your code can be altered to make use of std::vector:
void some_func(size_t n)
{
std::vector<int> data(n);
. . . // do stuff with the array
} // memory held by data will be freed here automatically
And if you are dealing with the memory in a local scope, consider using STLSoft's auto_buffer, which will allocate from an internal buffer (held on the stack, as part of the instance) if possible, only going to the heap if it cannot:
void some_func(size_t n)
{
stlsoft::auto_buffer<int, 10> data(n); // only allocates if n > 10
. . . // do stuff with the array
} // memory held by data will be freed here automatically, if any was allocated
Read more about auto_buffer.
The code as shown has the array either on the stack, or in initialized part of the data segment, i.e. you don't deallocate it (which, as mentioned by others, would be "undefined behavior".) Were it on the "free store", you'd do that with delete [] data.
Just as RichieHindle stated above when you want to free the space dynamically allocated for an array pointed by data you have to put two brackets [] between the reserved word delete and the pointer to the beginning of the allocated space. Since data can point to a single int in memory as well as to the first element in the array this is the only way you let the compiler know that you want to delete the whole chunk of memory. If you don't do it the proper way the behaviour is "undetermined" (Stroustrup, The C++ Programming Language).
Fixes C4154 and C4156 warnings
float AR[5] = { 1.0f, 2.0f, ..., ..., ...};
delete [] & AR;