x86_64: start the kernel bootstrap heap above 16MiB

this will keep the memory area below 16MiB free for DMA memory allocations.

arch/x86_64/init.c

@@ -18,7 +18,20 @@
 #include <kernel/types.h>
 #include <kernel/vm.h>
 
-#define PTR32(x) ((void *)((uintptr_t)(x)))
+#define PTR32(x)            ((void *)((uintptr_t)(x)))
+
+/* the physical address of the start of the memblock heap.
+ * this is an arbirary value; the heap can start anywhere in memory.
+ * any reserved areas of memory (the kernel, bsp, bios data, etc) are
+ * automatically taken into account.
+ * HOWEVER, this value will dictate how much physical memory is required for
+ * the kernel to boot successfully.
+ * the value of 16MiB (0x1000000) means that all heap allocations will be
+ * above 16MiB, leaving the area below free for DMA operations.
+ * this value CAN be reduced all the way to zero to minimise the amount of
+ * memory required to boot, but this may leave you with no DMA memory available.
+ */
+#define MEMBLOCK_HEAP_START 0x1000000
 
 static ml_cpu_block g_bootstrap_cpu = {0};
 
@@ -33,7 +46,7 @@ static void bootstrap_cpu_init(void)
 
 static void early_vm_init(uintptr_t reserve_end)
 {
-	uintptr_t alloc_start = VM_KERNEL_VOFFSET;
+	uintptr_t alloc_start = VM_KERNEL_VOFFSET + MEMBLOCK_HEAP_START;
 	/* boot code mapped 2 GiB of memory from
 	   VM_KERNEL_VOFFSET */
 	uintptr_t alloc_end = VM_KERNEL_VOFFSET + 0x7fffffff;