description

# allocate<T extends NativeType> abstract method

Pointer<T> allocate<T extends NativeType>(

1. int byteCount, {
2. int? alignment,

})

Allocates `byteCount` bytes of memory on the native heap.

If `alignment` is provided, the allocated memory will be at least aligned to `alignment` bytes.

To allocate a multiple of `sizeOf<T>()` bytes, call the allocator directly as a function: `allocator<T>(count)` (see AllocatorAlloc.call for details).

    // This allocates two bytes. If you intended two Int32's, this is an
    // error.
    allocator.allocate<Int32>(2);

    // This allocates eight bytes, which is enough space for two Int32's.
    // However, this is not the idiomatic way.
    allocator.allocate<Int32>(sizeOf<Int32>() * 2);

    // The idiomatic way to allocate space for two Int32 is to call the
    // allocator directly as a function.
    allocator<Int32>(2);

Throws an ArgumentError if the number of bytes or alignment cannot be satisfied.

## Implementation

    Pointer<T> allocate<T extends NativeType>(int byteCount, {int? alignment});