runtime: implement addrRanges.findSucc with a binary search

This change modifies addrRanges.findSucc to more efficiently find the
successor range in an addrRanges by using a binary search to narrow down
large addrRanges and iterate over no more than 8 addrRanges.

This change makes the runtime more robust against systems that may
aggressively randomize the address space mappings it gives the runtime
(e.g. Fuchsia).

For #40191.

Change-Id: If529df2abd2edb1b1496d8690ddd284ecd7138c2
Reviewed-on: https://go-review.googlesource.com/c/go/+/242679
Trust: Michael Knyszek <mknyszek@google.com>
Reviewed-by: Austin Clements <austin@google.com>
Reviewed-by: Michael Pratt <mpratt@google.com>

src/runtime/mranges.go

@@ -172,20 +172,46 @@ func (a *addrRanges) init(sysStat *sysMemStat) {
 	a.totalBytes = 0
 }
 
-// findSucc returns the first index in a such that base is
+// findSucc returns the first index in a such that addr is
 // less than the base of the addrRange at that index.
 func (a *addrRanges) findSucc(addr uintptr) int {
-	// TODO(mknyszek): Consider a binary search for large arrays.
-	// While iterating over these ranges is potentially expensive,
-	// the expected number of ranges is small, ideally just 1,
-	// since Go heaps are usually mostly contiguous.
 	base := offAddr{addr}
-	for i := range a.ranges {
+
+	// Narrow down the search space via a binary search
+	// for large addrRanges until we have at most iterMax
+	// candidates left.
+	const iterMax = 8
+	bot, top := 0, len(a.ranges)
+	for top-bot > iterMax {
+		i := ((top - bot) / 2) + bot
+		if a.ranges[i].contains(base.addr()) {
+			// a.ranges[i] contains base, so
+			// its successor is the next index.
+			return i + 1
+		}
+		if base.lessThan(a.ranges[i].base) {
+			// In this case i might actually be
+			// the successor, but we can't be sure
+			// until we check the ones before it.
+			top = i
+		} else {
+			// In this case we know base is
+			// greater than or equal to a.ranges[i].limit-1,
+			// so i is definitely not the successor.
+			// We already checked i, so pick the next
+			// one.
+			bot = i + 1
+		}
+	}
+	// There are top-bot candidates left, so
+	// iterate over them and find the first that
+	// base is strictly less than.
+	for i := bot; i < top; i++ {
 		if base.lessThan(a.ranges[i].base) {
 			return i
 		}
 	}
-	return len(a.ranges)
+	return top
 }
 
 // findAddrGreaterEqual returns the smallest address represented by a