【发布时间】:2012-09-03 07:19:15
【问题描述】:
我在控制台中遇到这样的错误:
:CGBitmapContextCreate:无效数据字节/行:对于 8 个整数位/分量、3 个分量、kCGImageAlphaPremultipliedFirst,应至少为 1920。 : CGBitmapContextCreateImage: 无效的上下文 0x0
我使用下面的代码:
- (UIImage *) imageFromSampleBuffer:(CMSampleBufferRef) sampleBuffer
{
// Get a CMSampleBuffer's Core Video image buffer for the media data
CVImageBufferRef imageBuffer = CMSampleBufferGetImageBuffer(sampleBuffer);
// Lock the base address of the pixel buffer
CVPixelBufferLockBaseAddress(imageBuffer, 0);
// Get the number of bytes per row for the pixel buffer
void *baseAddress = CVPixelBufferGetBaseAddress(imageBuffer);
// Get the number of bytes per row for the pixel buffer
size_t bytesPerRow = CVPixelBufferGetBytesPerRow(imageBuffer);
// Get the pixel buffer width and height
size_t width = CVPixelBufferGetWidth(imageBuffer);
size_t height = CVPixelBufferGetHeight(imageBuffer);
// Create a device-dependent RGB color space
CGColorSpaceRef colorSpace = CGColorSpaceCreateDeviceRGB();
// Create a bitmap graphics context with the sample buffer data
CGContextRef context = CGBitmapContextCreate(baseAddress, width, height, 8,
bytesPerRow, colorSpace, kCGBitmapByteOrder32Little | kCGImageAlphaPremultipliedFirst);
// Create a Quartz image from the pixel data in the bitmap graphics context
CGImageRef quartzImage = CGBitmapContextCreateImage(context);
// Unlock the pixel buffer
CVPixelBufferUnlockBaseAddress(imageBuffer,0);
// Free up the context and color space
CGContextRelease(context);
CGColorSpaceRelease(colorSpace);
// Create an image object from the Quartz image
UIImage *image = [UIImage imageWithCGImage:quartzImage];
// Release the Quartz image
CGImageRelease(quartzImage);
return (image);
}
【问题讨论】:
-
如果你NSLog
bytesPerRow、width和height,结果是什么? -
imageView中没有图片显示
-
我的意思是,如果你在
CGColorSpaceRef colorSpace = CGColorSpaceCreateDeviceRGB();之前加上NSLog(@"%zi, %zi, %zi",width,height,bytesPerRow);,那么控制台上的输出是什么? -
它给出这 3 个值 --480, --360, --724
-
那么
sampleBuffer出了点问题。有效的 BGRACMSampleBufferRef每行的字节数至少是所代表图像宽度的四倍。您的问题几乎可以肯定不在于这段代码(无论如何它只是从 Apple 复制粘贴的),而在于调用它的代码,从外观上看。
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