The question is as shown in the title. Thank you in advance for your answer! ：）

You divide the array size by the samplerate to get the number of seconds

@Balwyn Thank you so much. That's a really clear answer. I get it now!

@miyu I agree with @Balwyn, but if you just need a rough estimate, your thinking could have gone like this:

• notice that there are almost 6 cycles of that sinewave in that table, I'll guess it's 5.9
• at 262 cycles/second (i.e. frequency), there are 1/262 seconds/cycle (i.e. period)
• 5.9/262 = approx 0.0225 seconds

And at 44100 samples/second, that's approx 993 samples. Gosh, so close to 1000, I'd bet that table holds 1000 samples.

@jameslo It's an accurate and useful estimation method, thanks a lot!