Hi Damien, On Thu, 10 Jun 2010, Damien Zammit wrote: What do you mean by "the last jackframes%L samples" ?? The process() callback tells you to process nFrames samples. Are you wanting to access the samples the preceeded this? If so, you'll need to store them somewhere yourself. Jack doesn't store them. You don't necc. need to process them independently of the jack callback. However, you will need to create your own buffer to store your data. However, if your calculations can't be done within a single process() callback without taking too long, then you will either need to break up your calculation so that it will fit (time-wise) into the process() callback. If you can't do that, then you /will/ need to process them independently of the jack callback. -gabriel

Am Donnerstag, den 10.06.2010, 23:18 +1000 schrieb Damien Zammit: Hi here is the implementation I used for a time domaine convolution with a fixed filter kernel length inline void convolver_filter(float* input, float* output, int sf, unsigned int iconvolvefilter) { // select kernel filter if (iconvolvefilter >= sizeof(filters) / sizeof(filters[0])) { iconvolvefilter = 0; } // push last samples to the start for (int i=0; i < 45; i++) { result[i] = result[sf+i]; } // set the rest to zero for (int i=45; i < sf+46; i++) { result[i] = 0; } // Do convolution: for (int i = 0; i < sf; i++) { for (int j = 0; j < 45; j++) { result[i+j] += input[i] * filters[iconvolvefilter][j]; } } for (int i = 0; i < sf; i++) { *output++ = result[i]; } } result[] is a internal buffer with length = jack_buffer_size + kernel filter length. sf is the jack_buffer_size. greats hermann