Question

创建一个函数has_twenty_ones，如果游戏中至少有一个玩家有21，则返回true，否则返回false。此函数应使用twenty_ones函数。

function has_twenty_ones($game){
    function twenty_ones($game)
    {
        $players_with_score_21 = [];
        foreach ($game['players'] as $name => $player) {
            $distance = 21 - $player['score'];
            if ($distance < 0) {
                continue;
            }
            if ($distance == 21) {
                $players_with_score_21 = [$name];
            }
        }
        return $players_with_score_21;
    }
    return isset($players_with_score_21);

}

编码的最佳方法是什么

Answer 1

检查twenty_ones函数的返回是否为空，如果返回false，则返回false {0}返回twenty_ones。

function has_twenty_ones($game){
    function twenty_ones($game){
        $players_with_score_21 = [];
        foreach ($game['players'] as $name => $player) {
            $distance = 21 - $player['score'];
            if ($distance < 0) {
                continue;
            }
            if ($distance == 21) {
                $players_with_score_21 = [$name];
            }
        }
        return $players_with_score_21;
    }
    $playersWithScore = twenty_ones($game);
    if (!empty($playersWithScore)) {
       return $playersWithScore;
    } else {
      return false;
    } 
}

Answer 2

我不确定为什么你需要两个功能。

正如@RiggsFolly所提到的，你实际上并没有调用twenty_ones（）函数。为什么不拥有以下代码：

function has_twenty_ones($game)
{
    foreach($game['players'] as $name => $player)
    {
        $distance = 21 - $player['score'];
        if ($distance < 0) {
            continue;
        }
        // If at least one player has 21, return true.
        if($distance == 21) {
            return true;
        }
    }
    return false;
}

当遇到得分为21的玩家时，上述内容将返回true，否则将返回false。

Answer 3

function twenty_ones($game)
{
    $players_with_score_21 = [];
    foreach ($game['players'] as $name => $player) {
       $distance = 21 - $player['score'];
       if ($distance < 0) {
        continue;
       }
       if ($distance == 21) {
                    $players_with_score_21 = [$name];
       }
    }
    return $players_with_score_21;
}

function has_twenty_ones($game){
    if (count ($this->twenty_ones($game)) > 0 )
        return true;
    else 
       return false;
}

如何在函数内返回函数

3 个答案: