javascript - JS 递归函数破解

Question

我已经使用递归函数有一段时间了，我对当前的问题完全感到困惑。这是我的代码:

var setbackArray = new Array();
setbackArray = [5, 15, 20];
var positionArray = new Array();
positionArray = ["28.0", "28.0", "24.4", "24.4", "24.4", "28.0", "28.0", "28.0", "28.0", "28.0", "24.4", "28.0", "28.0", "28.0", "24.4", "24.4", "24.4", "24.4", "24.4", "24.4", "24.4", "24.4", "24.4", "24.4", "18.5", "18.5", "18.5", "18.5", "22.1", "22.1", "22.1", "22.1", "28.0", "28.0", "28.0", "28.0", "38.6", "38.6", "32.7", "32.7", "38.6", "32.7", "38.6", "32.7", "32.7", "38.6", "38.6", "38.6", "32.7", "32.7", "32.7", "38.6", "32.7", "38.6", "32.7", "38.6", "43.2", "43.2", "43.2", "43.2", "43.2", "43.2", "43.2", "43.2", "43.2", "43.2", "43.2", "43.2", "43.2", "43.2", "43.2", "43.2", "43.2", "43.2", "43.2", "43.2", "22.1", "22.1", "22.1", "22.1", "22.1", "22.1", "22.1", "22.1", "22.1", "22.1", "32.7", "32.7", "32.7", "32.7", "38.6", "38.6", "38.6", "38.6"]
var recursive = function (i, length) {
    console.log('i: ' + i + ', length: ' + length);
    if (i < length) {
        var seatposition = Number(positionArray[i]).toFixed(1);
        console.log(seatposition);
        if (seatposition < setbackArray[setbackArray.length - 1] + 20 && seatposition > setbackArray[0] - 20) {
            console.log('Doing Some Math.....');
        } else {
            console.log('Not Usable');
            recursive(++i, length);
        }
        console.log('Doing More Math.......');
        console.log('Doing Even More Math.......');
        console.log('Doing Last Bit Of Math.......');
        console.log('Display Stuff On Screen');
        recursive(++i, length);
    } else {
        console.log('done checking');
    }
}
recursive(0, positionArray.length);

在实际代码中，两个数组都是动态创建的，我只是在这里对它们进行编码，以便您有一个真实的示例。基本上我会遍历positionArray中的所有数字，看看该数字是否小于setbackArray中的最大数字加20并且大于setbackArray中的最小数字减20。如果是，我用它做一些数学运算稍后使用。如果不是，我希望它移动到positionArray中的下一个数字。

我遇到的问题是，一旦 i < length 不再为真，它会显示“完成检查”，然后将 i 重置为以前的值并继续运行。它会无休止地执行此操作并使整个页面崩溃。

我知道问题就在这里:

} else {
      console.log('Not Usable');
      recursive(++i, length);
}

如果我删除递归标注，它会正常运行，但会执行我不想对该数字执行的额外数学运算。

有什么想法吗？

工作示例 here

Answer 1

看来您在您确定的有问题的 else block 中的意图实际上是:

    } else {
        console.log('Not Usable');
        return recursive(++i, length);
    }

这会将函数的其余部分短路为“不可用”值。

就像现在一样，在这种情况下，您会递归调用函数两次，当递归沿着多个路径继续进行时，会导致控制流 fork 。

重构和修复该部分的另一种方法是:

    var seatposition = Number(positionArray[i]).toFixed(1);
    console.log(seatposition);
    if (seatposition < setbackArray[setbackArray.length - 1] + 20 && seatposition > setbackArray[0] - 20) {
        console.log('Doing Some Math.....');
        console.log('Doing More Math.......');
        console.log('Doing Even More Math.......');
        console.log('Doing Last Bit Of Math.......');
        console.log('Display Stuff On Screen');
    } else {
        console.log('Not Usable');
    }
    recursive(++i, length);