fix: 修正项目名称与输出目录的解析逻辑

examples/tetris-sdl/project.yml

@@ -5,8 +5,8 @@ sources:
   - main.php
   - ./php-src
   - ./cpp-src
-ldflags:
+ld-flags:
   - "-lSDL2"
-cxxflags:
+cxx-flags:
   - "-I/usr/include/SDL2"
   - "-D_REENTRANT"

phpunit/src/CompilerBaseApiTest.php

@@ -312,6 +312,21 @@ YAML, 'custom-name.yml', 'yaml-alias');
         $this->assertSame(dirname($projectFile) . '/out', $this->getPropertyValue('outputDir'));
     }
 
+    public function testParseProjectYamlNameDoesNotSetOutputDirectory(): void
+    {
+        $projectFile = $this->createProjectFile(<<<'YAML'
+name: tetris
+sources:
+  - main.php
+YAML, 'myproject.yml', 'examples/tetris-sdl');
+
+        $this->invokeMethod('parseProjectYaml', $projectFile);
+
+        $this->assertSame('tetris', $this->getPropertyValue('targetName'));
+        $this->assertSame('', $this->getPropertyValue('outputDir'));
+        $this->assertSame('tetris', $this->invokeMethod('getTargetFileName'));
+    }
+
     public function testParseProjectYamlResolvesRelativePathOptionsAgainstYamlDirectory(): void
     {
         $projectFile = $this->createProjectFile(<<<'YAML'

src/Php/Translator.php

@@ -2173,10 +2173,12 @@ CODE;
             }
         }
 
-        // 读取 output/name
-        $output = $cfg['output'] ?? $cfg['name'] ?? null;
+        // 读取 output/name。name 只表示目标名，不能按 YAML 目录解析成输出路径。
+        $output = $cfg['output'] ?? null;
         if (!empty($output)) {
             $this->setTargetName($this->resolvePath((string) $output, $projectDir, 'Output path'));
+        } elseif (!empty($cfg['name'])) {
+            $this->setTargetName((string) $cfg['name']);
         }
 
         // 读取 cpp-compiler