作者热门文章
- Java 双重比较
- java - 比较器与 Apache BeanComparator
- Objective-C 完成 block 导致额外的方法调用?
- database - RESTful URI 是否应该公开数据库主键?
我想从图像创建视频。
所以,我是引用链接的来源。
链接:CVPixelBufferPool Error ( kCVReturnInvalidArgument/-6661)
func writeAnimationToMovie(path: String, size: CGSize, animation: Animation) -> Bool {
var error: NSError?
let writer = AVAssetWriter(URL: NSURL(fileURLWithPath: path), fileType: AVFileTypeQuickTimeMovie, error: &error)
let videoSettings = [AVVideoCodecKey: AVVideoCodecH264, AVVideoWidthKey: size.width, AVVideoHeightKey: size.height]
let input = AVAssetWriterInput(mediaType: AVMediaTypeVideo, outputSettings: videoSettings)
let pixelBufferAdaptor = AVAssetWriterInputPixelBufferAdaptor(assetWriterInput: input, sourcePixelBufferAttributes: nil)
input.expectsMediaDataInRealTime = true
writer.addInput(input)
writer.startWriting()
writer.startSessionAtSourceTime(kCMTimeZero)
var buffer: CVPixelBufferRef
var frameCount = 0
for frame in animation.frames {
let rect = CGRectMake(0, 0, size.width, size.height)
let rectPtr = UnsafeMutablePointer<CGRect>.alloc(1)
rectPtr.memory = rect
buffer = pixelBufferFromCGImage(frame.image.CGImageForProposedRect(rectPtr, context: nil, hints: nil).takeUnretainedValue(), size)
var appendOk = false
var j = 0
while (!appendOk && j < 30) {
if pixelBufferAdaptor.assetWriterInput.readyForMoreMediaData {
let frameTime = CMTimeMake(Int64(frameCount), 10)
appendOk = pixelBufferAdaptor.appendPixelBuffer(buffer, withPresentationTime: frameTime)
// appendOk will always be false
NSThread.sleepForTimeInterval(0.05)
} else {
NSThread.sleepForTimeInterval(0.1)
}
j++
}
if (!appendOk) {
println("Doh, frame \(frame) at offset \(frameCount) failed to append")
}
}
input.markAsFinished()
writer.finishWritingWithCompletionHandler({
if writer.status == AVAssetWriterStatus.Failed {
println("oh noes, an error: \(writer.error.description)")
} else {
println("hrmmm, there should be a movie?")
}
})
return true;
}
func pixelBufferFromCGImage(image: CGImageRef, size: CGSize) -> CVPixelBufferRef {
let options = [
kCVPixelBufferCGImageCompatibilityKey: true,
kCVPixelBufferCGBitmapContextCompatibilityKey: true]
var pixBufferPointer = UnsafeMutablePointer<Unmanaged<CVPixelBuffer>?>.alloc(1)
let status = CVPixelBufferCreate(
nil,
UInt(size.width), UInt(size.height),
OSType(kCVPixelFormatType_32ARGB),
options,
pixBufferPointer)
CVPixelBufferLockBaseAddress(pixBufferPointer.memory?.takeUnretainedValue(), 0)
let rgbColorSpace = CGColorSpaceCreateDeviceRGB()
let bitmapinfo = CGBitmapInfo.fromRaw(CGImageAlphaInfo.NoneSkipFirst.toRaw())
var pixBufferData:UnsafeMutablePointer<(Void)> = CVPixelBufferGetBaseAddress(pixBufferPointer.memory?.takeUnretainedValue())
let context = CGBitmapContextCreate(
pixBufferData,
UInt(size.width), UInt(size.height),
8, UInt(4 * size.width),
rgbColorSpace, bitmapinfo!)
CGContextConcatCTM(context, CGAffineTransformMakeRotation(0))
CGContextDrawImage(
context,
CGRectMake(0, 0, CGFloat(CGImageGetWidth(image)), CGFloat(CGImageGetHeight(image))),
image)
CVPixelBufferUnlockBaseAddress(pixBufferPointer.memory?.takeUnretainedValue(), 0)
return pixBufferPointer.memory!.takeUnretainedValue()
有些东西甚至可以在电影之后像图像一样保留在内存中。我相信还是没有剩下 PixcelBuffer。
在 Objective-c 中我有一个方法 CVPixelBufferRelease(buffer) 来释放 PixcelBuffer,我不能再在 Swift 中使用它。我如何释放 PixcelBuffer 干什么?
如果有人能提供帮助,我将不胜感激。
1
2
最佳答案
当使用 CVPixelBufferCreate
时,UnsafeMutablePointer
必须在检索其内存
后被销毁。
当我创建一个 CVPixelBuffer
时,我是这样做的:
func allocPixelBuffer() -> CVPixelBuffer {
let pixelBufferAttributes : CFDictionary = [...]
let pixelBufferOut = UnsafeMutablePointer<CVPixelBuffer?>.alloc(1)
_ = CVPixelBufferCreate(kCFAllocatorDefault,
Int(Width),
Int(Height),
OSType(kCVPixelFormatType_32ARGB),
pixelBufferAttributes,
pixelBufferOut)
let pixelBuffer = pixelBufferOut.memory!
pixelBufferOut.destroy()
return pixelBuffer
}
关于ios - 我想快速释放 CVPixelBufferRef,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/29050062/
使用登录后,我想吐出用户名。 但是,当我尝试单击登录按钮时, 它给了我力量。 我看着logcat,但是什么也没显示。 这种编码是在说。 它将根据我在登录屏幕中输入的名称来烘烤用户名。 不会有任何密码。
关闭。这个问题不符合Stack Overflow guidelines .它目前不接受答案。 这个问题似乎是题外话,因为它缺乏足够的信息来诊断问题。 更详细地描述您的问题或include a min
我是一名优秀的程序员,十分优秀!