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我正在尝试在带有 xcode 的 Mac 上使用 glfw 2.7.5,但出现以下错误:

Undefined symbols for architecture i386:
  "_glfwGetKey", referenced from:
      Game::run() in Game.o
      RedScene::update(float, bool) in main.o
  "_glfwGetWindowParam", referenced from:
      Game::run() in Game.o
  "_glfwInit", referenced from:
      Window::Window() in Window.o
  "_glfwOpenWindow", referenced from:
      createWindow(int, int, int, int, int, int, int, int) in Window.o
  "_glfwOpenWindowHint", referenced from:
      Window::Window() in Window.o
  "_glfwPollEvents", referenced from:
      Game::run() in Game.o
  "_glfwSetKeyCallback", referenced from:
      createWindow(int, int, int, int, int, int, int, int) in Window.o
  "_glfwSetMouseButtonCallback", referenced from:
      createWindow(int, int, int, int, int, int, int, int) in Window.o
  "_glfwSetMousePosCallback", referenced from:
      createWindow(int, int, int, int, int, int, int, int) in Window.o
  "_glfwSetMouseWheelCallback", referenced from:
      createWindow(int, int, int, int, int, int, int, int) in Window.o
  "_glfwSetWindowTitle", referenced from:
      Window::setCaption(std::__1::basic_string<char, std::__1::char_traits<char>, std::__1::allocator<char> > const&) in Window.o
  "_glfwSwapBuffers", referenced from:
      RedScene::update(float, bool) in main.o
  "_glfwSwapInterval", referenced from:
      createWindow(int, int, int, int, int, int, int, int) in Window.o
  "_glfwTerminate", referenced from:
      Window::~Window() in Window.o
ld: symbol(s) not found for architecture i386
clang: error: linker command failed with exit code 1 (use -v to see invocation) 

我已经阅读了使用 GLFW 和 XCode 的问题:将无法编译以及互联网上的其他几个链接,但我找不到解决方案。

我已经设置了 Cocoa 框架、OpenGL 框架和 libglfw.a(我使用“sudo make cocoa-install”编译),在我设置 -lgflw 的链接器标志中,我正在为 32 位英特尔构建。起初我是为 64 位构建的,但我认为 GLFW 在 Mac 上还不支持它(我遇到了操纵杆文件错误),当我编译为 32 位时错误发生了变化,它找不到我的功能正在使用。知道我能做些什么来解决这个问题吗?

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1 回答 1

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很可能您已经为 64 位构建了 libglfw.a,因此当您尝试在 32 位应用程序中使用它时会出现链接错误。您可以为 32 位重建 libglfw.a(或作为“胖”32 位和 64 位库),或者您可以将您的应用程序构建为 64 位。

请注意,在最近的 Mac OS X 和 Xcode 版本中,gcc 默认构建 64 位代码,因此您需要-m32在命令行中指定才能获取 32 位代码。在实践中如何执行此操作取决于 libglfw.a 的生成文件 -ARCH您可以在命令行上指定一个或其他符号来强制执行 32 位构建。

于 2012-10-09T15:02:34.813 回答