Question

如何降低函数的圈复杂度，该函数返回依赖于3个布尔值的笛卡尔乘积的值？如何使以下代码看起来更干净？

这适用于学校项目，并不是作业本身的要求，但我通常会发现自己编写的函数依赖于 - 有时 - 相当复杂的真值表。我认为这不是最好的方法。

    public function getDiscount( $values ) {

    $res       = new stdClass();
    $res->code = 400;

    if ( ! is_bool( $values['new_customer'] ) || ! is_bool( $values['loyalty_card'] ) || ! is_bool( $values['coupon'] ) ) {
        $res->data = "Missing inputs";

        return $res;
    }

    if ( $values['new_customer'] == true && $values['loyalty_card'] == true && $values['coupon'] == true ) {
        $res->data = "Invalid input";

        return $res;
    }

    if ( $values['new_customer'] == true && $values['loyalty_card'] == true && $values['coupon'] == false ) {
        $res->data = "Invalid input";

        return $res;
    }

    $res->code = 200;

    if ( $values['new_customer'] == true && $values['loyalty_card'] == false && $values['coupon'] == true ) {
        $res->data = 20;

        return $res;
    }

    if ( $values['new_customer'] == true && $values['loyalty_card'] == false && $values['coupon'] == false ) {
        $res->data = 15;

        return $res;
    }

    if ( $values['new_customer'] == false && $values['loyalty_card'] == true && $values['coupon'] == true ) {
        $res->data = 30;

        return $res;
    }

    if ( $values['new_customer'] == false && $values['loyalty_card'] == true && $values['coupon'] == false ) {
        $res->data = 10;

        return $res;
    }

    if ( $values['new_customer'] == false && $values['loyalty_card'] == false && $values['coupon'] == true ) {
        $res->data = 20;

        return $res;
    }

    if ( $values['new_customer'] == false && $values['loyalty_card'] == false && $values['coupon'] == false ) {
        $res->data = 0;

        return $res;
    }

    $res->code = 400;
    $res->data = "Invalid input";
    return $res;

}

Answer 1

这是一种与语言无关的范例，您会发现不同的语言将具有不同的内置函数和运算符，以便于处理真值表。

My Favorite是C＃7.0切换语句https://visualstudiomagazine.com/articles/2017/02/01/pattern-matching.aspx
中的新模式匹配

首先，确定你的真相表

CASE    New Customer    Loyalty     Coupon  Output
  1        Yes            Yes        Yes    'Invalid Input'
  2        Yes            Yes        No     'Invalid Input'             
  3        Yes            No         Yes      20
  4        Yes            No         No       15                
  5        No             Yes        Yes      30
  6        No             Yes        No       10                
  7        No             No         Yes      20
  8        No             No         No       0

我们可以立即看到案例1和案例2不依赖于优惠券条件，因此这是您的第一次优化。要注意的第二个条件是，没有新客户是真实的并且他们有忠诚卡的情况，这是有道理的。从那里开始，在代码和程序中直观地遵循最简单的方法是使用嵌套分支，这样我们只评估每次条件检查一次。

public function getDiscount( $values ) {

    $res       = new stdClass();
    $res->code = 400;

    if ( ! is_bool( $values['new_customer'] ) || ! is_bool( $values['loyalty_card'] ) || ! is_bool( $values['coupon'] ) ) {
        $res->data = "Missing inputs";

        return $res;
    }

    if ( $values['new_customer'] == true && $values['loyalty_card'] == true ) {
        $res->data = "Invalid input";

        return $res;
    }

    $res->code = 200;

    // Check New Customer conditions
    if ( $values['new_customer'] == true) {
        if( $values['coupon'] == true ) 
            $res->data = 20;
        else
            $res->data = 15;
    }

    // Check existing customer conditions
    else {

        // Has Loyalty Card
        if ( $values['loyalty_card'] == true ) {
            // Has Coupon
            if( $values['coupon'] == true ) 
                $res->data = 30;
            else
                $res->data = 10;
        }

        // No Loyalty Card
        else {
            // Has Coupon
            if( $values['coupon'] == true ) 
                $res->data = 20;
            else
                $res->data = 0;
        }
    }

    // Don't need a default fail condition here, because we have covered all possible combinations     
    return $res;

}

使用值添加可能也更容易，真值表是表示下一代码块的好方法：

CASE    New Customer    Loyalty     Coupon  Output
  1        Yes            Yes        Yes    'Invalid Input'
  2        Yes            Yes        No     'Invalid Input'             
  3        Yes  +15       No         Yes +5   =20
  4        Yes  +15       No         No       =15               
  5        No             Yes +10    Yes +20  =30
  6        No             Yes +10    No       =10               
  7        No             No         Yes +20  =20
  8        No             No         No       0             

public function getDiscount( $values ) {

    $res       = new stdClass();
    $res->code = 400;

    if ( ! is_bool( $values['new_customer'] ) || ! is_bool( $values['loyalty_card'] ) || ! is_bool( $values['coupon'] ) ) {
        $res->data = "Missing inputs";

        return $res;
    }

    if ( $values['new_customer'] == true && $values['loyalty_card'] == true ) {
        $res->data = "Invalid input";

        return $res;
    }

    $res->code = 200;
    $res->data = 0;

    // Check New Customer conditions
    if ( $values['new_customer'] == true) {
        $res->data += 15;
        if( $values['coupon'] == true ) 
            $res->data += 5;
    }

    // Check existing customer conditions
    else {

        // Has Loyalty Card
        if ( $values['loyalty_card'] == true )
            $res->data += 10;

        // Has Coupon
        if( $values['coupon'] == true ) 
            $res->data += 20;
    }

    // Don't need a default fail condition here, because we have covered all possible combinations     
    return $res;

}

有许多方法可以使这只猫皮肤化:)上面的代码示例突出显示了如何使用分支逻辑仅评估每个条件一次。虽然在这种情况下微不足道，但评估未来的某些条件可能会产生严重的性能影响，因此您只需要评估一次。

如何降低函数的圈复杂度，该函数返回依赖于3个布尔值的笛卡尔乘积的值？

1 个答案: