How many cycles will cost when using a multiplicaiton?

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Just using * as the symbol of multiplication.

Does it depend on the data bit wide ?

Thanks

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The "*" operator will, by itself, produce a purely combinational multiplier. Wi

So, the question is not how many cycles it will take but how much delay it will add and, therefore, at which frequency can the design run.

And of course, the answer is that it will depend on the width.