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Make CMAKE_HOST_SYSTEM_NAME available in scripting context

Gregor Jasny 9 years ago
parent
commit
0bbd993f61

+ 0 - 2
Modules/CMakeDetermineSystem.cmake

@@ -35,7 +35,6 @@
 if(CMAKE_HOST_UNIX)
   find_program(CMAKE_UNAME uname /bin /usr/bin /usr/local/bin )
   if(CMAKE_UNAME)
-    exec_program(uname ARGS -s OUTPUT_VARIABLE CMAKE_HOST_SYSTEM_NAME)
     exec_program(uname ARGS -r OUTPUT_VARIABLE CMAKE_HOST_SYSTEM_VERSION)
     if(CMAKE_HOST_SYSTEM_NAME MATCHES "Linux|CYGWIN.*|Darwin|^GNU$")
       exec_program(uname ARGS -m OUTPUT_VARIABLE CMAKE_HOST_SYSTEM_PROCESSOR
@@ -67,7 +66,6 @@ if(CMAKE_HOST_UNIX)
   endif()
 else()
   if(CMAKE_HOST_WIN32)
-    set (CMAKE_HOST_SYSTEM_NAME "Windows")
     if (DEFINED ENV{PROCESSOR_ARCHITEW6432})
       set (CMAKE_HOST_SYSTEM_PROCESSOR "$ENV{PROCESSOR_ARCHITEW6432}")
     else()

+ 10 - 0
Source/cmStateSnapshot.cxx

@@ -18,6 +18,10 @@
 #include "cmVersion.h"
 #include "cmake.h"
 
+#if !defined(_WIN32)
+#include <sys/utsname.h>
+#endif
+
 #if defined(__CYGWIN__)
 #include "cmSystemTools.h"
 #endif
@@ -298,9 +302,15 @@ void cmStateSnapshot::SetDefaultDefinitions()
 #if defined(_WIN32)
   this->SetDefinition("WIN32", "1");
   this->SetDefinition("CMAKE_HOST_WIN32", "1");
+  this->SetDefinition("CMAKE_HOST_SYSTEM_NAME", "Windows");
 #else
   this->SetDefinition("UNIX", "1");
   this->SetDefinition("CMAKE_HOST_UNIX", "1");
+
+  struct utsname uts_name;
+  if (uname(&uts_name) == 0) {
+    this->SetDefinition("CMAKE_HOST_SYSTEM_NAME", uts_name.sysname);
+  }
 #endif
 #if defined(__CYGWIN__)
   std::string legacy;

+ 0 - 1
Tests/CMakeTests/ToolchainTest.cmake.in

@@ -48,7 +48,6 @@ set(CMAKE_SYSTEM_NAME)
 set(CMAKE_SYSTEM_VERSION)
 set(CMAKE_SYSTEM_PROCESSOR)
 set(CMAKE_HOST_SYSTEM)
-set(CMAKE_HOST_SYSTEM_NAME)
 set(CMAKE_HOST_SYSTEM_VERSION)
 set(CMAKE_HOST_SYSTEM_PROCESSOR)