prog/alloc: align address allocation for aligned[addr]

Calls to alloc didn't respect the alignment attribute. Now
Type.Alignment() is used to ensure each type is correctly
aligned. Existing descriptions with [align[X]] don't have an
issue as they align to small blocks and default align is to
64 bytes. This commits adds support for [align[X]] for an X
larger than 64.

1 files changed, 8 insertions, 3 deletions

diff --git a/prog/alloc.go b/prog/alloc.go
index 344ec7a0e..a4b7b7f1f 100644
--- a/prog/alloc.go
+++ b/prog/alloc.go
@@ -51,13 +51,18 @@ func (ma *memAlloc) noteAlloc(addr0, size0 uint64) {
 	}
 }
 
-func (ma *memAlloc) alloc(r *randGen, size0 uint64) uint64 {
+// alloc returns the next free address of size0 with respect to the given alignment.
+func (ma *memAlloc) alloc(r *randGen, size0, alignment0 uint64) uint64 {
 	if size0 == 0 {
 		size0 = 1
 	}
+	if alignment0 == 0 {
+		alignment0 = 1
+	}
 	size := (size0 + memAllocGranule - 1) / memAllocGranule
+	alignment := (alignment0 + memAllocGranule - 1) / memAllocGranule
 	end := ma.size - size
-	for start := uint64(0); start <= end; start++ {
+	for start := uint64(0); start <= end; start += alignment {
 		empty := true
 		for i := uint64(0); i < size; i++ {
 			if ma.get(start + i) {
@@ -72,7 +77,7 @@ func (ma *memAlloc) alloc(r *randGen, size0 uint64) uint64 {
 		}
 	}
 	ma.bankruptcy()
-	return ma.alloc(r, size0)
+	return ma.alloc(r, size0, alignment0)
 }
 
 func (ma *memAlloc) bankruptcy() {