Support run or test target invoked within debugger

Useful, as it gets all the environment produced by the makefile. For the
test target, it might make sense to have the child process debug plugin
for VS installed.

win32/build/Makefile

@@ -29,6 +29,14 @@ MCFILE=$(BUILD_DIR)\wsyslog.rc
 BUILD_DIR_DEV_NAME=php-$(PHP_VERSION_STRING)-devel-$(PHP_COMPILER_SHORT)-$(PHP_ARCHITECTURE)
 BUILD_DIR_DEV=$(BUILD_DIR)\$(BUILD_DIR_DEV_NAME)
 
+!if "$(DEBUGGER)" == "1"
+DEBUGGER_CMD=devenv
+DEBUGGER_ARGS=/debugexe
+!else
+DEBUGGER_CMD=
+DEBUGGER_ARGS=
+!endif
+
 all: generated_files $(EXT_TARGETS) $(PECL_TARGETS) $(SAPI_TARGETS)
 
 build_dirs: $(BUILD_DIR) $(BUILD_DIRS_SUB) $(BUILD_DIR_DEV)
@@ -151,18 +159,19 @@ clean-pgo: clean-all
 	-del /f /q $(BUILD_DIR)\pecl-$(PHP_VERSION_STRING)$(PHP_ZTS_ARCHIVE_POSTFIX)-Win32-$(PHP_COMPILER_SHORT)-$(PHP_ARCHITECTURE).zip
 	-del /f /q $(BUILD_DIR)\php-test-pack-$(PHP_VERSION_STRING).zip
 
+
 !if $(PHP_TEST_INI_PATH) == ""
 test: set-tmp-env
-	"$(BUILD_DIR)\php.exe" -d open_basedir= -d output_buffering=0 run-tests.php $(TESTS) -p "$(BUILD_DIR)\php.exe"
+	$(DEBUGGER_CMD) $(DEBUGGER_ARGS) "$(BUILD_DIR)\php.exe" -d open_basedir= -d output_buffering=0 run-tests.php $(TESTS) -p "$(BUILD_DIR)\php.exe"
 
 run: set-tmp-env
-	"$(BUILD_DIR)\php.exe" $(ARGS)
+	$(DEBUGGER_CMD) $(DEBUGGER_ARGS) "$(BUILD_DIR)\php.exe" $(ARGS)
 !else
 test: set-tmp-env
-	"$(BUILD_DIR)\php.exe" -n -c $(PHP_TEST_INI_PATH) -d open_basedir= -d output_buffering=0 -d memory_limit=-1 run-tests.php -p "$(BUILD_DIR)\php.exe" -n -c $(PHP_TEST_INI_PATH) $(TESTS)
+	$(DEBUGGER_CMD) $(DEBUGGER_ARGS) "$(BUILD_DIR)\php.exe" -n -c $(PHP_TEST_INI_PATH) -d open_basedir= -d output_buffering=0 -d memory_limit=-1 run-tests.php -p "$(BUILD_DIR)\php.exe" -n -c $(PHP_TEST_INI_PATH) $(TESTS)
 
 run: set-tmp-env
-	"$(BUILD_DIR)\php.exe" -n -c $(PHP_TEST_INI_PATH) $(ARGS)
+	$(DEBUGGER_CMD) $(DEBUGGER_ARGS) "$(BUILD_DIR)\php.exe" -n -c $(PHP_TEST_INI_PATH) $(ARGS)
 !endif
 
 build-snap: set-tmp-env generated_files