If you like the original tweet, I will CC you when I post the answer. See this for more information on LP64: https://en.wikipedia.org/wiki/64-bit_computing#64-bit_data_models …pic.twitter.com/4HC5npWfFy
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If you like the original tweet, I will CC you when I post the answer. See this for more information on LP64: https://en.wikipedia.org/wiki/64-bit_computing#64-bit_data_models …pic.twitter.com/4HC5npWfFy
First of all big h/t to @Myriachan whom I learned this devious example from
The answer is D, this is undefined behavior but how you might protest, unsigned number don’t have undefined behavior on overflow, do they?
This is an unfortunate result of the Usual Arithmetic Conversions: https://en.cppreference.com/w/c/language/conversion#Usual_arithmetic_conversions … Before performing binary operations we need to bring the operands to a common type via the Usual Arithmetic Conversions. See [expr.mul]p2 http://eel.is/c++draft/expr.mul#2 …pic.twitter.com/oWGx6V6fol
Hi @shafikyaghmour
I was not in CC of your last quizz
if you can make it , that's cool
thanks a lot
Apologies, I do it by hand so I must have messed up last time 
I know this issue and it really bugs me, i'll keep quiet why until later. But it makes mepic.twitter.com/KKC1klo967
No regrets!
Operands are promoted to int, preserving value; signed multiplication on int then overflows, causing UB.
Hi, why are operands promoted to int?
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