Discussion
Left shifting an unsigned int or char by 1 is always equivalent to multiplying it by 2.
Answer: A
Correct answer: True
For unsigned types the arithmetic is defined to wrap modulo \( 2^n \), so a left shift by one is exactly a multiplication by two with any overflow discarded rather than undefined.
That is the whole difference from the signed case, where shifting a value whose significant bits reach the sign position is undefined behaviour. The char in the question is promoted to int before the shift, so the operation happens in a wider type; the guarantee is what makes shift-based multiplication safe on unsigned values and unsafe on signed ones.
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