-# Should __WXDEBUG__ be defined? The default value "default" means that it will
-# be defined if BUILD=debug and not defined if BUILD=release. [0,1,default]
-DEBUG_FLAG = default
+# What type of library to build? [0,1]
+SHARED = 0
+
+# Build wxUniversal instead of native port? [0,1]
+WXUNIV = 0
+
+# Compile Unicode build of wxWidgets? [0,1]
+UNICODE = 0
+
+# Use MSLU library when building Unicode version. [0,1]
+MSLU = 0
+
+# Type of compiled binaries [debug,release]
+BUILD = debug
+
+# The target processor architecture must be specified when it is not X86.
+# This does not affect the compiler output, so you still need to make sure
+# your environment is set up appropriately with the correct compiler in the
+# PATH. Rather it affects some options passed to some of the common build
+# utilities such as the resource compiler and the linker. [AMD64,IA64]
+TARGET_CPU = $(CPU)