allocate<T extends NativeType> method
Null safety
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});