8:24 the implicit binary point is before 1 and not between 1 and 0 in (1010) it is 1.Mentissa where “1.” Is always implied. But the rest is explained very well thanks
@Vishal no that's not correct. he is iterating through a couple of models to represent the floating point numbers. and in that specific model you mentioned, he is using the model where the leading 1 in mantissa (significand) is explicit, NOT implicit. if you follow through, in the next iteration he shows the model where you can use implicit 1. that's a teaching style I guess. Start with the inefficient model and walk your way up to more efficient ones. that way you understand the topic more easily, and appreciate the progress we made.
Don't forget the implicit extra bit. Double's significand precision is really 53 bits (check wiki or the IEEE 754 paper), where 52 bits are stored explicitly and the implicit bit is considered 1 for nonzero exponents.
Thank you Kris! An exceptionally good explanation of floating-point numbers. All of the pressing questions in my mind have been answered :). I am looking forward to using your channel as a source for further learning.
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8:24 the implicit binary point is before 1 and not between 1 and 0 in (1010) it is 1.Mentissa where “1.” Is always implied. But the rest is explained very well thanks
That is very true which makes the result wrong as well. I've got 6.5 as a result. Do you agree?
@Vishal no that's not correct. he is iterating through a couple of models to represent the floating point numbers. and in that specific model you mentioned, he is using the model where the leading 1 in mantissa (significand) is explicit, NOT implicit. if you follow through, in the next iteration he shows the model where you can use implicit 1. that's a teaching style I guess. Start with the inefficient model and walk your way up to more efficient ones. that way you understand the topic more easily, and appreciate the progress we made.
He knows that and explains it later on in the video. See the big "Proposal" in the upper left corner.
Slight error @ 29.48. Double precision significant bits is 52 not 53- You're missing the sign bit :)
Don't forget the implicit extra bit. Double's significand precision is really 53 bits (check wiki or the IEEE 754 paper), where 52 bits are stored explicitly and the implicit bit is considered 1 for nonzero exponents.
Great video, exactly what I was looking for, subscribed 👍
i m not related to computer science but this is awsome
very clear explain video 👍 help me a lot understanding IEEE754
Thank you Kris! An exceptionally good explanation of floating-point numbers. All of the pressing questions in my mind have been answered :). I am looking forward to using your channel as a source for further learning.
Great explanation! There is an error in 31:49: The number of significand bits of 64-bit double should be 52 instead of 53:)
No, it's correct. The implicit leading bit (explained around 25:00) is also part of the significand.
Magnificent!
Thank you so much for explaining minifloat crystal clear.
thank you so much .... i was stuck on this for 4 hours
Thank you so much!!!
can anyone solve this -9/5 to binary using 8-bit floating point notation
very helpful video i searching for two days
YOU SAVED ME THANK YOU THIS WAS SUCH A GREAT VIDEO
Outstanding Sir! Thx
Thank you very much for such a instructive video!
Excellent
Encode the following values into the floating-point format
a) 2 3/4 b) -3 1/2
please
2 3/4 = 0 100 0110
(−1)^0×(10)^(100 − 011)×1.0110
-3 1/2 = 1 100 1100
(−1)^1×(10)^(100 − 011)×1.1100
i can't understand the significant value...
Why is this different to what I learned at university?
I imagine you learned full with IEEE floating points? The only meaningful difference should be the bit width.