我在内存中有一个完整的视频文件,我想用libav来解码整个帧.我该怎么办?关键是我可以使用avformat_open_input()函数直接从文件中读取,但我确实需要从存储在内存中的文件中进行.
我的AVIOContext实现:
class AVIOMemContext
{
public:
AVIOMemContext(char* videoData, const int videoLen)
{
// Output buffer
bufferSize = 32768;
buffer = static_cast<char*>(av_malloc(bufferSize));
// Internal buffer
pos = 0;
this->videoData = videoData;
this->videoLen = videoLen;
ctx_ = avio_alloc_context((unsigned char*) buffer, bufferSize, AVIO_FLAG_READ, this, &AVIOMemContext::read, &AVIOMemContext::write, &AVIOMemContext::seek);
}
~AVIOMemContext()
{
av_free(videoData);
}
static int read(void *opaque, unsigned char *buf, int buf_size)
{
AVIOMemContext* This = static_cast<AVIOMemContext*>(opaque);
// Read from pos to pos + buf_size
if (This->pos + buf_size > This->videoLen)
{
int len …
Run Code Online (Sandbox Code Playgroud) 我不在乎next()
或error()
重视。我是否仍需要将空函数定义为函数中的回调subscribe()
?
我在使用Google Maps API V2时遇到了一些问题.我已经尝试了很多教程并搜索了很多答案(包括堆栈溢出),但我找到的所有都没有用.我可以用xml标签绘制地图<fragment>
.没问题,它有效.但是当我试图让地图MainActivity.java
getMap()
返回null时我不知道为什么.
MainActivity.java
package com.example.testmaps;
import android.app.Activity;
import android.os.Bundle;
import com.google.android.gms.maps.GoogleMap;
import com.google.android.gms.maps.MapFragment;
import com.google.android.gms.maps.model.LatLng;
import com.google.android.gms.maps.model.MarkerOptions;
public class MainActivity extends Activity {
private GoogleMap mMap;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
mMap = ((MapFragment) getFragmentManager().findFragmentById(R.id.map)).getMap();
//HERE mMap IS NULL
mMap.addMarker(new MarkerOptions()
.position(new LatLng(10, 10))
.title("Hello world"));
}
}
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activity_main.xml中
<RelativeLayout xmlns:android="http://schemas.android.com/apk/res/android"
xmlns:tools="http://schemas.android.com/tools"
android:layout_width="match_parent"
android:layout_height="match_parent"
tools:context=".MainActivity" >
<fragment
android:id="@+id/map"
android:layout_width="match_parent"
android:layout_height="match_parent"
android:name="com.google.android.gms.maps.MapFragment"/>
</RelativeLayout>
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的Manifest.xml
<?xml version="1.0" encoding="utf-8"?>
<manifest xmlns:android="http://schemas.android.com/apk/res/android"
package="com.example.testmaps"
android:versionCode="1" …
Run Code Online (Sandbox Code Playgroud) 我陷入了zend框架,我是新手,试图实现一个网站只是为了学习它.所以我的网站是关于比萨饼的.当我尝试添加披萨时,在发送表单数据后,我收到此错误消息说:致命错误:在C:\ wamp\www\pizzalast\module\PizzaPoint\src\PizzaPoint中找不到类'PizzaPoint\Controller\Pizza'第27行的Controller\PizzaController.php,在这一行中,我实际上是一个位于"Model"文件夹中的类Pizza的对象.
这是披萨控制器的添加动作:
public function addAction()
{
$form = new PizzaForm();
$form->get('submit')->setValue('Add');
$request = $this->getRequest();
if($request->isPost())
{
$pizza = new Pizza();
$form->setInputFilter($pizza->getInputFilter());
$form->setData($request->getPost());
if($form->isValid())
{
$pizza->exchangeArray($form->getData());
$this->getPizzaTable()->savePizza($pizza);
}
}
return array('form' => $form);
}
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这些是Pizza.php文件的前40行代码:
namespace PizzaPoint\Model;
use Zend\InputFilter\InputFilter;
use Zend\InputFilter\InputFilterAwareInterface;
use Zend\InputFilter\InputFilterInterface;
class Pizza implements InputFilterAwareInterface {
public $id;
public $title;
public $zutaten;
public $smallprice;
public $bigprice;
public $familyprice;
public $partyprice;
protected $inputFilter;
public function exchangeArray($data)
{
$this->id = (!empty($data['id'])) ? $data['id'] : null;
$this->title = (!empty($data['title'])) ? $data['title'] …
Run Code Online (Sandbox Code Playgroud) 我有一个数组:
my @y = keys(%{$table->{fields}});
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我想得到该数组的引用:
$columns = \@y;
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有没有办法在一行上执行此操作?
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