How to initialise memory with new operator in C++?

It's a surprisingly little-known feature of C++ (as evidenced by the fact that no-one has given this as an answer yet), but it actually has special syntax for value-initializing an array:

new int[10]();

Note that you must use the empty parentheses — you cannot, for example, use (0) or anything else (which is why this is only useful for value initialization).

This is explicitly permitted by ISO C++03 5.3.4[expr.new]/15, which says:

A new-expression that creates an object of type T initializes that object as follows:

...

• If the new-initializer is of the form (), the item is value-initialized (8.5);

and does not restrict the types for which this is allowed, whereas the (expression-list) form is explicitly restricted by further rules in the same section such that it does not allow array types.

There is number of methods to allocate an array of intrinsic type and all of these method are correct, though which one to choose, depends...

Manual initialisation of all elements in loop

int* p = new int[10];
for (int i = 0; i < 10; i++)
    p[i] = 0;

Using std::memset function from <cstring>

int* p = new int[10];
std::memset(p, 0, sizeof *p * 10);

Using std::fill_n algorithm from <algorithm>

int* p = new int[10];
std::fill_n(p, 10, 0);

Using std::vector container

std::vector<int> v(10); // elements zero'ed

If C++11 is available, using initializer list features

int a[] = { 1, 2, 3 }; // 3-element static size array
vector<int> v = { 1, 2, 3 }; // 3-element array but vector is resizeable in runtime

Assuming that you really do want an array and not a std::vector, the "C++ way" would be this

#include <algorithm> 

int* array = new int[n]; // Assuming "n" is a pre-existing variable

std::fill_n(array, n, 0); 

But be aware that under the hood this is still actually just a loop that assigns each element to 0 (there's really not another way to do it, barring a special architecture with hardware-level support).