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Dear All,
could you kindly help me with deleting this std::map. Here are some details...
typedef std::map<std::string,int> N2v;
N2v n2v;
//Some looping construct
n2v.insert(....)
//End of some looping construct
//Now for the delete :
for(int i=0;i<n2v.size();i++){
// std::map<int>.swap(n2v[i]); -> IS WRONG. KINDLY GUIDE ME IN FIXING IT
}
std::map<std::string,int>().swap(n2v);
Thanks,
Ravi
What do you mean by "deleting"?
If it is something like "erasingall the elements of a map" then use clear().
See map Members
What's wrong with just letting the destructor of the map do its job?Quote:
Originally Posted by sunshekar
The N2v map is deleted once the end of the local block is reached. There is no need to do any deletions yourself.Code:typedef std::map<std::string,int> N2v;
void foo()
{
// some code
{
N2v n2v;
//End of some looping construct
} // N2v is now destroyed
}
Regards,
Paul McKenzie
Why are you using swap()? You appear to be trying to use the "swap trick", but that is for vectors. With a map, you can just call .clear().
Hi all,
some clarifications:
1. By deleting, I mean removing from memory. We have observed that a simple clear does not release the memory for 2D vectors (unlike 1D vectors) , even when it goes out of scope. Hence the need for this delete i.e. releasing it from memory.
2. "Why are you using swap()? You appear to be trying to use the "swap trick", but that is for vectors. With a map, you can just call .clear().": Are you sure? In 2D vectors, the memory need not be released at the end of scope (unlike 1D vectors).
Thanks,
Ravi
Could you describe the way you "have observed that a simple clear does not release the memory for 2D vectors (unlike 1D vectors) , even when it goes out of scope"?Quote:
Originally Posted by sunshekar
And please, define "does not release the memory".
You are using a std::map, not vectors.Quote:
Originally Posted by sunshekar
And all of that coding could be a big waste of time. All you're doing by coding "delete loops" is just wasting cycles that accomplish nothing.Quote:
Are you sure? In 2D vectors, the memory need not be released at the end of scope (unlike 1D vectors).
Once the map goes out of scope, anything else after that is dependent on the heap manager and OS. The map has done its job by deallocating its internal structure. After that the heap manager is responsible to see if the memory is returned back to the OS, or keep it for later allocations -- it isn't up to you to decide this or even the map class (unless you code a custom allocator and call Win32 heap functions to allocate/deallocate memory).
The question is this: why your application needs to do any of this?
Regards.
Paul McKenzie
For instance,
inside a function....
int Nproc = 2000;
std::vector<std::vector<double>> test(Nproc);
for(int i=0;i<Nproc;i++){
for(int j=0;j<1000;j++){
test[i].push_back( (j*1.0));
}
}
//End of function
The compilers nowadays may not deallocate this test vector immediately even after it goes afer scope. They are sort of "optimised" to dellocate them in some unknown way. This can be verified in large production type codes, by using the "top" command in linux. One can see that the memory is not automatically deallocated even after the routine is over (for 2D vectors).
By releasing the memory, I mean deallocating the memory.
Warm Regards,
Ravi
How are you defining a 2D vector? If it's a vector<vector<T>>, then the destructors of the inner vector objects ought to be called when the outer vector is cleared (even if the memory they're sitting in isn't released), unless you're using a vector class which has some kind of special optimization that keeps them around.Quote:
Originally Posted by sunshekar
Defined as given in my earlier post. All the entities are doubles, not any special class.
Nevertheless, based on our experiences doing this swap() is the only way we can be certain that the memory is deallocated.
Warm Regards,
Ravi
You need to use code tags when posting code.Quote:
Originally Posted by sunshekar
See how that looks?Code:int Nproc = 2000;
std::vector<std::vector<double>> test(Nproc);
for(int i=0;i<Nproc;i++){
for(int j=0;j<1000;j++){
test[i].push_back( (j*1.0));
}
}
Now:
The compiler does deallocate memory, else you have a memory leak. What you want to do goes way beyond what the compiler is doing.Quote:
The compilers nowadays may not deallocate this test vector immediately even after it goes afer scope.
Read my previous post. Control of whether deallocated memory goes back to the OS is not the job of the compiler. That job goes to the CRT heap manager.
The vector class isn't magic. It isn't doing anything that isn't C++. A vector class uses the same new/delete calls as your own code would do. So the issue you're seeing is not a vector or a map issue. It is an issue with the heap manager and how it controls memory.
Regards,
Paul McKenzie
All the swap() does is make sure the memory is released while the vector is still in scope. If you have two vectors in scope, and one of them is not needed anymore, then the swap trick has merit.Quote:
Originally Posted by sunshekar
Once that vector goes out of scope, then the "go out of scope" becomes the ultimate "swap trick", as any and all calls to delete the vector's contents are issued via the vector's destructor.
Again, a vector isn't magic -- it isn't coded with another language that somehow knows to call the OS to return the memory, and cannot do things beyond what the C++ language provides. A vector is no different than any class you would write. It is calling new, delete, etc. just like you would call new and delete. Yes, a vector may be complex, but still, it's just good old C++.
Regards,
Paul McKenzie