我有一个32位的wav数组,我想把它转换为8位
所以我尝试将此功能转换为32到16
void _waveIn_DataAvailable(object sender, WaveInEventArgs e)
{
byte[] newArray16Bit = new byte[e.BytesRecorded / 2];
short two;
float value;
for (int i = 0, j = 0; i < e.BytesRecorded; i += 4, j += 2)
{
value = (BitConverter.ToSingle(e.Buffer, i));
two = (short)(value * short.MaxValue);
newArray16Bit[j] = (byte)(two & 0xFF);
newArray16Bit[j + 1] = (byte)((two >> 8) & 0xFF);
}
}
并修改它以x位为目的地
private byte[] Convert32BitRateToNewBitRate(byte[] bytes, int newBitRate)
{
var sourceBitRate = 32;
byte[] newArray = new byte[bytes.Length / (sourceBitRate / newBitRate)];
for (int i = 0, j = 0; i < bytes.Length; i += (sourceBitRate / 8), j += (newBitRate / 8))
{
var value = (BitConverter.ToSingle(bytes, i));
var two = (short)(value * short.MaxValue);
newArray[j] = (byte)(two & 0xFF);
newArray[j + 1] = (byte)((two >> 8) & 0xFF);
}
return newArray;
}
我的问题是,我不确定如何转换&#34; for&#34;中的代码。循环,我试图调试它,但我无法弄清楚它是如何工作的。
我在这里看到:simple wav 16-bit / 8-bit converter source code?,他们将值除以256得到16到8,我试图除以256得到32到16但它没有工作
for (int i = 0, j = 0; i < bytes.Length; i += sourceBitRateBytes, j += newBitRateBytes)
{
var value = BitConverter.ToInt32(bytes, i);
value /= (int)Math.Pow(256, sourceBitRate / newBitRate / 2.0);
var valueBytes = BitConverter.GetBytes(value);
for (int k = 0; k < newBitRateBytes; k++)
{
newArray[k + j] = valueBytes[k];
}
答案 0 :(得分:0)
for循环在以下位置仍然使用16位:
short.MaxValue
。请改用byte.MaxValue
。[j]
和[j+1]
分配两个字节。仅分配一个字节。我没有剩下的程序,没有样本数据,所以我很难尝试。但我会说下面关于正确的声音
for (int i = 0, j = 0; i < bytes.Length; i += (sourceBitRate / 8), j += (newBitRate / 8))
{
var value = (BitConverter.ToSingle(bytes, i));
var two = (byte)(value * byte.MaxValue);
newArray[j] = two;
}
请注意,这仅适用于8位,因此newBitRate
必须为8,否则无效。它应该不是该方法的参数。