Thank you very much. Very nice, clear, and right-to-the-point explanation. This and your other videos helped me a lot in my computer architecture class. I'm going to look into your other videos/playlists. Thanks again.
What happens when the most significant bit of the multiplicand is 1? The moment you shift it to the left, you lose the value and your product is scrap.
That's why the register for the multiplicand is twice the size of the largest value it's designed to handle. For a 32-bit architecture, the register for the multiplicand will be 64 bits, allowing the number to be shifted left 32 times (once for each bit in the 32-bit multiplier) without losing any data. For a 64-bit architecture, the register for the multiplicand is 128 bits.
Thank you so much, even 6 years later this video is helping CS students who didn't get great professors.
I love how clear these explanations are. Much love from Egypt.
Thank you Padraic for your great lectures.
Regards,
Yakov.
Thank you very much. Very nice, clear, and right-to-the-point explanation. This and your other videos helped me a lot in my computer architecture class. I'm going to look into your other videos/playlists. Thanks again.
thankyou so much. where do you teach? i love your explanation
your videos are GREAT!!
Very Clear !!! Thank you professor~
this is very clear explanation
Thanks for the clear explanation!!
You didnt consider carry out in the shift right operation, so thats going to be a problem for some numbers. Try multiplying binary equivalent of 14x13
could you please do signed multiplication hardware?Thank you
Great tutor
does anyone know some book that can be useful?
Made my concepts clear
very well explained ty sir
What happens when the most significant bit of the multiplicand is 1? The moment you shift it to the left, you lose the value and your product is scrap.
That's why the register for the multiplicand is twice the size of the largest value it's designed to handle. For a 32-bit architecture, the register for the multiplicand will be 64 bits, allowing the number to be shifted left 32 times (once for each bit in the 32-bit multiplier) without losing any data. For a 64-bit architecture, the register for the multiplicand is 128 bits.
I would favorite this video if I could. Very clear.
why dont we just stop when the multiplier register becomes 0?
Isn't the second addition (ignored one) wrong?
Nice catch!
Great explanation
THANK YOU :)))))))))))))))))))))))))))))))))))
Excellent!
ty man
Thank you
thx alot
you seem very blocked up prof! anyways, great explanation
this dude got A L L E R G I E S