Question

我需要重复将1024+连续4字节浮点数（范围-1到1）转换为2字节短路（范围-32768到32767）并写入磁盘。

目前我通过循环执行此操作：

short v = 0;
for (unsigned int sample = 0; sample < length; sample++) 
{
    v = (short)(inbuffer[sample * 2] * 32767.0f);
    fwrite(&v, 2, 1, file);
}

这可行，但浮点计算和循环很昂贵。有什么办法可以优化吗？

Answer 1

short v = 0;
for (unsigned int sample = 0; sample < length; sample++) 
{
    v = (short)(inbuffer[sample * 2] * 32767.0f);
    // The problem is not here-------^^^^^^^^^^^
    fwrite(&v, 2, 1, file);        
    // it is here ^^^^^^^
}

典型的Mac（Objective-c标签，或者我们在这里谈论iphone？）每秒可以进行数十亿次浮点运算。然而，fwrite是一个库调用，它跟随一些间接，将其数据写入某个缓冲区并可能刷新它。最好是批量填充自己的缓冲区：

short v[SZ] = 0; // make sure SZ is always > length, or allocate a working buffer on the heap. for (unsigned int sample = 0; sample < length; sample++) { v[sample] = (short)(inbuffer[sample * 2] * 32767.0f); } fwrite(v,sizeof(v),1,file);

Answer 2

我原本以为重复拨打fwrite会是一个昂贵的部分。怎么样：

short outbuffer[length]; // note: you'll have to malloc this if length isn't constant and you're not using a version of C that supports dynamic arrays.
for (unsigned int sample = 0; sample < length; sample++) 
{
    outbuffer[sample] = (short)(inbuffer[sample * 2] * 32767.0f);
}
fwrite(outbuffer, sizeof *outbuffer, length, file);

Answer 3

我想，你的循环的瓶颈可能不是浮动转换的短暂，而是将输出写入文件 - 尝试将文件输出移到循环之外

short v = 0;
short outbuffer = // create outbuffer of required size
for (unsigned int sample = 0; sample < length; sample++) 
{
    outbuffer[sample] = (short)(inbuffer[sample * 2] * 32767.0f);
}

fwrite(outbuffer, 2, sizeof(outbuffer), file);

Answer 4

您可以尝试这样的事情：

out[i] = table[((uint32_t *)in)[i]>>16];

其中table是一个查找表，它将IEEE浮点的高16位映射到所需的int16_t值。然而，这将失去一些精确性。您需要保留并使用23位（1个符号位，8个指数位和14个尾数位）以获得完全精度，这意味着一个16 MB的表，这将破坏缓存一致性，从而消除性能。

您确定浮点转换速度慢吗？只要你以fwrite方式使用fwrite，你就会在int16_t上花费50到100倍的浮点运算时间。如果你处理这个问题并且代码仍然太慢，你可以使用一种方法来添加一个魔术偏差并读取尾数位以转换为{{1}}而不是乘以32767.0。这可能会或可能不会更快。

将范围-1的浮点数转换为1的快速方法是短的？

4 个答案: