>... The time it takes > logically is zero: The full block is written into the delay line > immediatly (that is: at the same time, delread~ reads it, see below).
OK, But how can a full block be written to a delay line before it has happened? You say the time it takes is logically zero, but in order to write a full block of samples to a delayline, 64 bits of the input signal would have to have elapsed before this can happen surely? For example: lets say there is an input signal to the explicitly defined sequence of delwrite~ - delread~ - output At t=0 the sample starts playing. Surely only until t=64 can a block be written by delwrite~ (immediately), and read (immediately) by delread~? I think I am thinking about this in the wrong way. Sorry for being slow :p K _______________________________________________ [email protected] mailing list UNSUBSCRIBE and account-management -> http://lists.puredata.info/listinfo/pd-list
