Let $$\\mathbf{h}$$ be a vector of size $$P$$ containing the impulse response you wish to convolve with, and let $$\\mathbf{x}$$ contain an unknown number of float values which is the input audio stream.<\/p>\n
Note that an FIR filter can be expressed as follows:
\n$$FIR(x[n],\\mathbf{h})=y[n]=\\sum_{j=0}^{P-1}({x[n+j] \\cdot h[j]})$$<\/p>\n
By chopping up $$x[n]$$ into sections of length $$L$$, we can apply the overlap-save algorithm to calculate the output of the filter.<\/p>\n
Let:
\n$$x[n]=\\sum_{r=0}^\\infty x_r[n-rL]$$<\/p>\n
where:
\n$$x_r[n]=x[n+r(L-P+1)-P+1] \\qquad 0\\leq n\\leq L-1$$ and 0 otherwise.
\nWe have each $$x_r$$ overlapping the previous one by $$P-1$$ points.<\/p>\n
Next we convolve each $$x_r[n]$$ with $$h[n]$$ and discard $$0\\leq n \\leq P-2$$ of the output $$y[n]$$ where:<\/p>\n
$$y[n]=x_r[n]*h[n]$$ (where $$*$$ refers to fourier convolution)<\/p>\n
Discard $$0\\leq n \\leq P-2$$ of $$y[n]$$, ie $$P-1\\leq n \\leq L-1$$ are the valid outputs for $$y[n]$$.<\/p>\n
This does not solve the problem of buffer management, you still have to work out how to manage the samples coming in if they come in blocks of size $$K$$ from an audio stream. <\/p>\n