Question

每当我抓住一段我正在处理的代码时，我都会得到This function's cyclomatic complexity is too high. (7)。但我对如何以这种方式重写它有点困惑，所以它有效。

这将是不断抛出该消息的功能：

function () {
  var duration = +new Date() - start.time,
    isPastHalf = Number(duration) < 250 && Math.abs(delta.x) > 20 || Math.abs(delta.x) > viewport / 2,
    direction = delta.x < 0;

  if (!isScrolling) {
    if (isPastHalf) {
      if (direction) {
        this.close();
      } else {
        if (this.content.getBoundingClientRect().left > viewport / 2 && pulled === true) {
          this.close();
          return;
        }
        this.open();
      }
    } else {
      if (this.content.getBoundingClientRect().left > viewport / 2) {
        if (this.isEmpty(delta) || delta.x > 0) {
          this.close();
          return;
        }
        this.open();
        return;
      }
      this.close();
    }
  }
}

我想听听一些关于如何以这种方式构建代码的建议，以避免这种情况。

Answer 1

你的代码中只有两个动作，但条件太多了。在条件中使用单个if-else语句和布尔运算符。如果这是不可能的，你至少可以

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这是你的功能简化：

var duration = +new Date() - start.time,
    isPastHalf = Number(duration) < 250 && Math.abs(delta.x) > 20 || Math.abs(delta.x) > viewport / 2,
    isFarRight = this.content.getBoundingClientRect().left > viewport / 2, 
    direction = delta.x < 0;

if (!isScrolling) {
    if (isPastHalf) {
        if (direction)
            this.close();
        else {
            if (isFarRight && pulled)
                this.close();
            else
                this.open();
        }
    } else {
        if (isFarRight) {
            // Looks like the opposite of `direction`, is it?
            if (this.isEmpty(delta) || delta.x > 0)
                this.close();
            else
                this.open();
        } else
            this.close();
    }
}

并缩短：

var duration = +new Date() - start.time,
    isPastHalf = Number(duration) < 250 && Math.abs(delta.x) > 20 || Math.abs(delta.x) > viewport / 2,
    isFarRight = this.content.getBoundingClientRect().left > viewport / 2, 
    direction = delta.x < 0,
    undirection = this.isEmpty(delta) || delta.x > 0;

if (!isScrolling) {
    if ( isPastHalf && !  direction && !(isFarRight && pulled)
     || !isPastHalf && !undirection &&  isFarRight )
        this.open();
    else
        this.close();
}

Answer 2

实际上所有这些return陈述都令人困惑，但它们提供了解决方案的提示。

if (direction) {
  this.close();
} else {
  if (this.content.getBoundingClientRect().left > viewport / 2 && pulled === true) {
    this.close();
    return; // We'll never `this.open()` if this is true anyway, so combine the booleans.
  }
  this.open();
}

怎么样：

if (direction || (this.content.getBoundingClientRect().left > viewport / 2 && pulled === true)) {
  this.close();
} else {
  this.open();
}

至于：

if (this.content.getBoundingClientRect().left > viewport / 2) {
  if (this.isEmpty(delta) || delta.x > 0) {
    this.close();
    return; // Combine the booleans!
  }
  this.open();
  return;
}

简化：

if ((this.isEmpty(delta) || delta.x > 0) || !this.content.getBoundingClientRect().left > viewport / 2) {
  this.close();
} else {
  this.open();
}

（旁白：原来的帖子遗漏了一个右大括号。如果你（OP）打算让这个功能继续超过你的帖子，那么这个答案是错的（但你应该更清楚了））

结果：我们已经取消了两项（重复的）决定：

function () {
  var duration = +new Date() - start.time,
    isPastHalf = Number(duration) < 250 && Math.abs(delta.x) > 20 || Math.abs(delta.x) > viewport / 2,
    direction = delta.x < 0;

  if (!isScrolling) {
    if (isPastHalf) {
      if (direction || (this.content.getBoundingClientRect().left > viewport / 2 && pulled === true)) {
        this.close();
      } else {
        this.open();
      }
    } else {
      if ((this.isEmpty(delta) || delta.x > 0) || !this.content.getBoundingClientRect().left > viewport / 2) {
        this.close();
      } else {
        this.open();
      }
    }
  }
}

Answer 3

首先，您的功能可以有三个结果：什么都不做，请致电this.close()或致电this.open()。因此理想情况下，结果函数将只有一个if语句来确定使用哪个结果。

下一步是将所有布尔代码提取到变量中。例如var leftPastCenter = this.content.getBoundingClientRect().left > viewport / 2。

最后，使用布尔逻辑逐步简化它。

我是这样做的：

首先，提取所有布尔变量：

function () {
    var duration = +new Date() - start.time,
      isPastHalf = Number(duration) < 250 && Math.abs(delta.x) > 20 || Math.abs(delta.x) > viewport / 2,
      direction = delta.x < 0,
      leftPastCenter = this.content.getBoundingClientRect().left > viewport / 2,
      positiveDelta = this.isEmpty(delta) || delta.x > 0,
      isPulled = pulled === true; // I'll assume the test is needed rather than just using pulled.

    if (!isScrolling) {
        if (isPastHalf) {
            if (direction) {
                this.close();
            } else {
                if (leftPastCenter && isPulled) {
                    this.close();
                    return;
                }
                this.open();
            }
        } else {
            if (leftPastCenter) {
                if (positiveDelta) {
                    this.close();
                    return;
                }
                this.open();
                return;
            }
            this.close();
        }
    }
}

最简单的部分就是要意识到isScrolling是否属实，什么都不会发生。这立即摆脱了一级嵌套：

    // above same
    if (isScrolling) { return; }

    if (isPastHalf) {
        if (direction) {
            this.close();
        } else {
            if (leftPastCenter && isPulled) {
                this.close();
                return;
            }
            this.open();
        }
    } else {
        if (leftPastCenter) {
            if (positiveDelta) {
                this.close();
                return;
            }
            this.open();
            return;
        }
        this.close();
    }
}

现在看一下调用this.open()的案例。如果isPastHalf为真，则仅在this.open()和!direction时调用!(leftPastCenter && isPulled)。如果isPastHalf为false，则仅在this.open()和leftPastCenter时调用!positiveDelta：

    // above same
    if (isScrolling) { return; }

    if (isPastHalf) {
        if (!direction && !(leftPastCenter && isPulled)) {
            this.open();
        } else {
            this.close();
        }
    } else {
        if (leftPastCenter && !positiveDelta) {
            this.open();
        } else {
            this.close();
        }
    }

翻转ifs（首先是this.close()），使代码更整洁，并给出我的最终版本：

    function () {

    var duration = +new Date() - start.time,
      isPastHalf = Number(duration) < 250 && Math.abs(delta.x) > 20 || Math.abs(delta.x) > viewport / 2,
      direction = delta.x < 0,
      leftPastCenter = this.content.getBoundingClientRect().left > viewport / 2,
      positiveDelta = this.isEmpty(delta) || delta.x > 0,
      isPulled = pulled === true; // I'll assume the test is needed rather than just using pulled.

    if (isScrolling) { return; }

    if (isPastHalf) {
        if (direction || (leftPastCenter && isPulled)) {
            this.close();
        } else {
            this.open();
        }
    } else {
        if (!leftPastCenter || positiveDelta) {
            this.close();
        } else {
            this.open();
        }
    }
}

在不知道您的代码库的情况下，我很难做得更多。需要注意的一点是direction，我的新变量positiveDelta几乎相同 - 您可以删除positiveDelta并使用direction。另外，direction不是布尔值的好名字，像movingLeft这样的东西会更好。

Answer 4

Bergi已经给出了正确答案，但它仍然太复杂，不符合我的口味。由于我们不是using fortran77，我认为我们最好使用early return。此外，可以通过引入额外的变量来进一步阐明代码：

function doSomething(isScrolling, start, delta, viewport) {
    if (isScrolling) return;

    var duration = +new Date() - start.time;
    var isPastHalf = Number(duration) < 250 && Math.abs(delta.x) > 20 || Math.abs(delta.x) > viewport / 2;
    var isFarRight = this.content.getBoundingClientRect().left > viewport / 2;

    // I'm not sure if my variable names reflect the actual case, but that's
    // exactly the point. By choosing the correct variable names for this,
    // anybody reading the code can immediatly comprehend what's happening.
    var isMovingToLeft = delta.x < 0;
    var isMovedPastEnd = isPastHalf && !isMovingToLeft && !(isFarRight && pulled);
    var isMovedBeforeStart = !isPastHalf && isMovingToLeft && isFarRight;

    if (isMovedPastEnd || isMovedBeforeStart) {
        this.open();
    else
        this.close();
    }
}

Answer 5

我更喜欢一个简单且嵌套较少的代码，如下所示：

function() 
{
    var duration = +new Date() - start.time,
        isPastHalf = Number(duration) < 250 && Math.abs(delta.x) > 20 || Math.abs(delta.x) > viewport / 2,
        direction = delta.x < 0;

    if (isScrolling)
    {
        return;
    }
    if (isPastHalf) 
    {
        if (direction) 
        {
            this.close();
            return;
        }
        if (this.content.getBoundingClientRect().left > viewport / 2 && pulled == = true) 
        {
            this.close();
            return;
        }
        this.open();
        return;
    }
    if (this.content.getBoundingClientRect().left > viewport / 2) 
    {
        if (this.isEmpty(delta) || delta.x > 0) 
        {
            this.close();
            return;
        }
        this.open();
        return;
    }
    this.close();
    return;
}

我怎么能减少圈复杂度？

5 个答案: