Iterative Implementation Double Iterativepower Double Base Int Exponent Double Retval 10 E Q43853441
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Iterative implementation double iterativePower(double base, int exponent){ double retVal 1.0; if (exponent < 0){ return 1.0 / iterativePower (base, -exponent); }else{ for (int i=0; i<exponent; i++) retVal *= base; } return retVal; } Recursive implementation double recursivePower(double base, int exponent){ if (exponent < 0){ return 1.0 / recursivePower(base, -exponent); 0){ }else if (exponent return 1.0; }else { return base * recursivePower(base, exponent 1); } } Show transcribed image text Iterative implementation double iterativePower(double base, int exponent){ double retVal 1.0; if (exponent
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