Question

鉴于此递归函数将base的数字提高为exponent的幂：

function power(base, exponent) {
  if (exponent == 0)
    return 1;
  else  return base * power(base, exponent - 1);
}

在内存中以什么顺序处理调用堆栈。

Answer 1

假设您用$charges = [ $row->StandardConcretePanel * $standardConcretePanelCharge, $row->ADAPedestrianRampWithDomes * 1200, $row->ADAPedestrianRampNoDomes * 800, $row->OversizedConcretePanel * 340, $row->ThickConcretePanel8to10 * 320, $row->ThickConcretePanel10to12 * 340, $row->ThickConcretePanel12orMore * 360, $row->HighEarlyConcretePanels * 1100, $row->WinterMix * 75, $row->ConcreteBlanketInstallation * 50, $row->PanelRemovalFillWithClassV * $panelRemovalCharge, $row->BarricadedAndProtectedPanels * 20, $row->HighProfilePanelsFrontOfBusinesses * 1100, $row->BeforeOrAfterHoursPanels * 1500, $row->AlleyOrDrivewayPanelWithIntegralCurb * 1200, $row->ConcreteShortLoads4PanelsOrLess * 400, $row->CurbAndGutter * 1200 ];调用它等于两个或更多，最终exponent将为零，并且函数的最后一次调用将返回exponent。然后倒数第二个调用将返回1，即base * 1。那么倒数第二个呼叫将返回base（这是base * base * 1的平方），依此类推，直到原始呼叫将返回您请求的功率为止。

Answer 2

这是调用该函数的示例：

 power (2,3) // base =2 , exponent = 3

    first call => (exponent is 3) !=0 // if is not true

        return base * power (2 , 2) //exponent now equals 2

        second Call => (exponent is 2) != 0 

            return base * power(2,1) // exponent now equals 1

            third call => (exponent is 1) != 0

                    return base * power(2,0) // exponent now equals 0

                        fourth call => (exponent is 0) == 0

                            return 1;

now replace the variables with its values
fourth = 1
third = base * fourth
second = base * third
first  = base * second

third = 2 * 1
second = 2 * 2
first = 2 * 4

递归调用按什么顺序堆叠

2 个答案: