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GPUImage构建错误
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Stack Overflow用户
提问于 2013-11-02 16:45:27
回答 1查看 917关注 0票数 0

我正在尝试使用GPUImage,已经成功地添加了目标和构建,但是当我在viewController和构建中使用它时,有一些问题我不能理解。有人知道为什么吗?非常感谢。

代码:

代码语言:javascript
复制
UIImage *inputImage = [UIImage imageNamed:@"1"];

GPUImagePicture *stillImageSource = [[GPUImagePicture alloc] initWithImage:inputImage];
GPUImageSepiaFilter *stillImageFilter = [[GPUImageSepiaFilter alloc] init];

[stillImageSource addTarget:stillImageFilter];
[stillImageSource processImage];

UIImage *currentFilteredVideoFrame = [stillImageFilter imageFromCurrentlyProcessedOutput];


UIImageView *image = [[UIImageView alloc] initWithFrame:CGRectMake(60, 60, 200, 300)];
image.image = currentFilteredVideoFrame;
[self.view addSubview:image];
[image release];

错误:

代码语言:javascript
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ld: warning: ignoring file /Users/xiaodong/Library/Developer/Xcode/DerivedData/SilentGif-cuagrzcstwdoemeamjiufacggbfa/Build/Products/Debug-iphonesimulator/libGPUImage.a, file was built for archive which is not the architecture being linked (x86_64): /Users/xiaodong/Library/Developer/Xcode/DerivedData/SilentGif-cuagrzcstwdoemeamjiufacggbfa/Build/Products/Debug-iphonesimulator/libGPUImage.a
Undefined symbols for architecture x86_64:
  "_OBJC_CLASS_$_GPUImageSepiaFilter", referenced from:
      objc-class-ref in ThirdViewController-5FAAC33C6210C63D.o
ld: symbol(s) not found for architecture x86_64
clang: error: linker command failed with exit code 1 (use -v to see invocation)
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回答 1

Stack Overflow用户

发布于 2014-02-28 13:20:27

您可以在GPUImage.xcodeproj中将ArchitectureStandard architectures更改为Standard architectures (including 64-bit),如下所示。

这是一个功能问题,请参阅this

票数 0
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页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/19740323

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