How do I create an array in C++ which is on the he

问题:

I have a very large array which must be 262144 elements in length (and potentially much larger in future). I have tried allocating the array on the stack like so:

#define SIZE 262144
int myArray[SIZE];

However, it appears that when I try and add elements past a certain point, the values are different when I try to access them. I understand that this is because there is only a finite amount of memory on the stack, as opposed to the heap which has more memory.

I have tried the following without much luck (does not compile):

#define SIZE 262144
int *myArray[SIZE] = new int[SIZE];

And then I considered using malloc, but I was wondering if there was a more C++ like way of doing this...

#define SIZE 262144
int *myArray = (int*)malloc(sizeof(int) * SIZE);

Should I just go with malloc?

回答1:

You'll want to use new like such:

int *myArray = new int[SIZE];

I'll also mention the other side of this, just in case....

Since your transitioning from the stack to the heap, you'll also need to clean this memory up when you're done with it. On the stack, the memory will automatically cleanup, but on the heap, you'll need to delete it, and since its an array, you should use:

delete [] myArray;

回答2:

The more C++ way of doing it is to use vector. Then you don't have to worry about deleting the memory when you are done; vector will do it for you.

#include <vector>

std::vector<int> v(262144);

回答3:

As the number is not necessarily known at compile time, the type is a pointer:

int *myArray = new int[262144];

回答4:

new allocates on the heap.

#define SIZE 262144
int * myArray = new int[SIZE];

回答5:

I believe

#define SIZE 262144
int *myArray[SIZE] = new int[262144];

should be

#define SIZE 262144
int *myArray = new int[262144];

or better yet

#define SIZE 262144
int *myArray = new int[SIZE];

回答6:

The reason the first try didn't work is that the syntax is incorrect. Try this:

int *myArray = new int[SIZE];

回答7:

Your syntax for using new was wrong, it should be:

int *myArray = new int[262144];

you only need to put the size on the right of the assignment.

However, if you're using C++ you might want to look at using std::vector (which you will have) or something like boost::scoped_array to make the the memory management a bit easier.