<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[yaveymao的专栏]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/mywfool</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; mywfool]]></copyright><item><title><![CDATA[使用mp4v2将aac音频h264视频数据封装成mp4开发心得]]></title><link>https://blog.youkuaiyun.com/mywfool/article/details/8581663</link><guid>https://blog.youkuaiyun.com/mywfool/article/details/8581663</guid><author>mywfool</author><pubDate>Fri, 15 Feb 2013 18:44:47 +0800</pubDate><description><![CDATA[这阵子在捣鼓一个将游戏视频打包成本地可播放文件的模块。开始使用avi作为容器，弄了半天无奈avi对aac的支持实在有限，在播放时音视频时无法完美同步。
关于这点avi文档中有提到：
For AAC, one RAW AAC frame usually spans over 1024 samples. However, depending on
the source container (e.]]></description><category></category></item><item><title><![CDATA[使用汇编优化BCB6的itoa]]></title><link>https://blog.youkuaiyun.com/mywfool/article/details/2295101</link><guid>https://blog.youkuaiyun.com/mywfool/article/details/2295101</guid><author>mywfool</author><pubDate>Tue, 15 Apr 2008 20:41:00 +0800</pubDate><description><![CDATA[今天写了个测试itoa性能的程序，做了1千万次itoa操作，主要代码如下：
char buf[64];
unsigned long t = GetTickCount();
for (int i = 0; i 
    itoa(i, buf, 10);
}
unsigned long t2 = GetTickCount();
printf("last=%s, %u/n", buf,]]></description><category></category></item></channel></rss>