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Javascript 如何优化接收数据?_Javascript_Google Chrome - Fatal编程技术网

Javascript 如何优化接收数据?

Javascript 如何优化接收数据?,javascript,google-chrome,Javascript,Google Chrome,我正在读取串行数据,它能更优化ab2str转换器吗 var lineBuffer = ""; var ab2str = function(buf) { var bufView = new Uint8Array(buf); var encodedString = String.fromCharCode.apply(null, bufView); return decodeURIComponent(escape(encodedString)); }; function(receiveI

我正在读取串行数据,它能更优化ab2str转换器吗

var lineBuffer = "";
var ab2str = function(buf) {
  var bufView = new Uint8Array(buf);
  var encodedString = String.fromCharCode.apply(null, bufView);
  return decodeURIComponent(escape(encodedString));
};

function(receiveInfo) {
    var ch = ab2str(receiveInfo.data);
    ch = ch.toString('utf8');
    switch (ch) {
      case "\u0002":
        console.log('>>> STARTED');
        break;      
      case "\u0003":            
        console.log(lineBuffer, lineBuffer.length);
        qrcode = lineBuffer;
        lineBuffer = "";
        console.log('>>> ENDED');
        break;
      default:
        //console.log('>>> ADDING in TROLL: ' , ab2str(receiveInfo.data));
        lineBuffer += ab2str(receiveInfo.data);
        break;
    }   
}

什么是
功能(receiveInfo)
?据我所知,它应该抛出一个意外的令牌异常,不是吗?它正在工作还是你只是在寻找一个优化?是的-它的工作序列。onReceive。addListener(函数(receiveInfo){},但我想对它进行更多的优化。