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where ‘op’ is one of ‘+’, ‘*’ and ‘ ’ (exponentiation) can be evaluated on a CPU with single register without  storing the value of (a * b) if
  • a)
     ‘op’ is ‘+’ or ‘*’
  • b)
     ‘op’ is ‘ ’ or ‘*’
  • c)
     ‘op’ is ‘ ’ or ‘+’
  • d)
    not possible to evaluate without storing
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
where ‘op’ is one of ‘+’, ‘*’ and ...
↑ has higer precedence than {*,+,-,/}
So, if op = ↑ implies, we need to evaluate the right hand side of ↑ first and then do the lhs part, which would definately require us to store the value of lhs 
but if its a '+' or '*' , we dont need to store the values evaluated, and on the go can do the operation directly on one register.
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Most Upvoted Answer
where ‘op’ is one of ‘+’, ‘*’ and ...
↑ has higer precedence than {*,+,-,/}
So, if op = ↑ implies, we need to evaluate the right hand side of ↑ first and then do the lhs part, which would definately require us to store the value of lhs 
but if its a '+' or '*' , we dont need to store the values evaluated, and on the go can do the operation directly on one register.
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Community Answer
where ‘op’ is one of ‘+’, ‘*’ and ...
↑ has higer precedence than {*,+,-,/}
So, if op = ↑ implies, we need to evaluate the right hand side of ↑ first and then do the lhs part, which would definately require us to store the value of lhs 
but if its a '+' or '*' , we dont need to store the values evaluated, and on the go can do the operation directly on one register.
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