allocate<T extends NativeType> 抽象方法

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

在本地堆上分配 `byteCount` 字节的内存。

如果提供了 `alignment`,则分配的内存至少将对齐到 `alignment` 字节数。

为了分配 `sizeOf<T>()` 字节的多倍,请直接将分配器作为函数调用:`allocator<T>(count)`(有关详细信息,请参阅 AllocatorAlloc.call)。

// 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);

如果字节数或对齐无法满足,将抛出 `ArgumentError`。

实现

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