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[Qemu-devel] [PATCH 8/9] linux-user: Default to 64k guest base

From: Alexander Graf
Subject: [Qemu-devel] [PATCH 8/9] linux-user: Default to 64k guest base
Date: Sat, 6 Jul 2013 02:36:17 +0200

Most kernels these days have protection code in place to forbid user space
to access low memory. The barrier varies between architectures though.

For this purpose we have the guest base option that allows us to offset
guest visible memory from host memory, so that the guest process thinks
it can access lower memory than it really can access.

Set the default for the guest base to 64k which should be good enough on
any host system.

This fixes running i386 wine on ARM for me.

Signed-off-by: Alexander Graf <address@hidden>
 linux-user/main.c |    6 +++---
 1 files changed, 3 insertions(+), 3 deletions(-)

diff --git a/linux-user/main.c b/linux-user/main.c
index 7f15d3d..a246cff 100644
--- a/linux-user/main.c
+++ b/linux-user/main.c
@@ -45,8 +45,8 @@ envlist_t *envlist;
 const char *cpu_model;
 unsigned long mmap_min_addr;
-unsigned long guest_base;
-int have_guest_base;
+unsigned long guest_base = 64 * 1024;
+int have_guest_base = 1;
 #if (TARGET_LONG_BITS == 32) && (HOST_LONG_BITS == 64)
  * When running 32-on-64 we should make sure we can fit all of the possible
@@ -3294,7 +3294,7 @@ static void handle_arg_cpu(const char *arg)
 static void handle_arg_guest_base(const char *arg)
     guest_base = strtol(arg, NULL, 0);
-    have_guest_base = 1;
+    have_guest_base = guest_base ? 1 : 0;
 static void handle_arg_reserved_va(const char *arg)

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