Javascript 如何避免复杂布尔表达式上的三值化
在linting我的Javascript时,我在我的一个复杂的三元选项上遇到了一个Javascript 如何避免复杂布尔表达式上的三值化,javascript,boolean,ternary-operator,Javascript,Boolean,Ternary Operator,在linting我的Javascript时,我在我的一个复杂的三元选项上遇到了一个没有不必要的三元警告 我知道如何在简单的布尔表达式上解决这个问题: var obvious = (1 === 1) ? true : false; // can simply become: var obvious = (1 === 1); 然而,在下面的布尔表达式中,我不知道如何适当缩小这个范围,而不必担心打破一些非常复杂的东西: const include = (options.directory &am
没有不必要的三元
警告
我知道如何在简单的布尔表达式上解决这个问题:
var obvious = (1 === 1) ? true : false;
// can simply become:
var obvious = (1 === 1);
然而,在下面的布尔表达式中,我不知道如何适当缩小这个范围,而不必担心打破一些非常复杂的东西:
const include =
(options.directory && file !== '.') ? false :
(!dotted) ? true :
(dotted && options.all) ? true :
(dotted && !implied && options.almostall) ? true :
(options.directory && file === '.') ? true :
false;
什么是正确的速记实现
试一试:
这是否正确?当您使用一组链式三元运算符编写代码时,代码会变得更简洁,通常可读性较差
const include =
(options.directory && file !== '.') ? false :
(!dotted) ? true :
(dotted && options.all) ? true :
(dotted && !implied && options.almostall) ? true :
(options.directory && file === '.') ? true :
false;
要对此进行分解,我将首先使用模块模式展开它:
include = (function () {
//set up some simple names for concepts:
var directory = options.directory;
var isDot = file === '.';
var all = options.all;
var almost = options.almostall;
if (directory && !isDot)
return false;
if (!dotted)
return true;
if (dotted && all)
return true;
if (dotted && implied && almost)
return true;
if (directory && isDot)
return true;
return false;
}());
这可以简化。检查后!虚线
,虚线
必须为真,并成为冗余:
转换为:
a
return a;
return a || b;
return !a && b;
!a || !b
作为一件足够好的事情,您可以随意停在这里,知道代码是简单和高效的
当然……这可以简化。最后一个
if
语句可以更改为返回值
:
转换为:
a
return a;
return a || b;
return !a && b;
!a || !b
当然,可以通过将最后一个if
转换为return
来简化:
转换为:
a
return a;
return a || b;
return !a && b;
!a || !b
……而且:
include = (function () {
//set up some simple names for concepts:
var directory = options.directory;
var isDot = file === '.';
var all = options.all;
var almost = options.almostall;
if (directory && !isDot)
return false;
if (!dotted)
return true;
return (all) ||
(implied && almost) ||
(directory && isDot);
}());
include = (function () {
//set up some simple names for concepts:
var directory = options.directory;
var isDot = file === '.';
var all = options.all;
var almost = options.almostall;
if (directory && !isDot)
return false;
return (!dotted) ||
(all) ||
(implied && almost) ||
(directory && isDot);
}());
……而且:
include = (function () {
//set up some simple names for concepts:
var directory = options.directory;
var isDot = file === '.';
var all = options.all;
var almost = options.almostall;
if (directory && !isDot)
return false;
if (!dotted)
return true;
return (all) ||
(implied && almost) ||
(directory && isDot);
}());
include = (function () {
//set up some simple names for concepts:
var directory = options.directory;
var isDot = file === '.';
var all = options.all;
var almost = options.almostall;
if (directory && !isDot)
return false;
return (!dotted) ||
(all) ||
(implied && almost) ||
(directory && isDot);
}());
……而且:
include = (function () {
//set up some simple names for concepts:
var directory = options.directory;
var isDot = file === '.';
var all = options.all;
var almost = options.almostall;
if (directory && !isDot)
return false;
if (!dotted)
return true;
return (all) ||
(implied && almost) ||
(directory && isDot);
}());
include = (function () {
//set up some simple names for concepts:
var directory = options.directory;
var isDot = file === '.';
var all = options.all;
var almost = options.almostall;
if (directory && !isDot)
return false;
return (!dotted) ||
(all) ||
(implied && almost) ||
(directory && isDot);
}());
转换为:
a
return a;
return a || b;
return !a && b;
!a || !b
这可以通过使用以下方法进一步简化:
转换为:
a
return a;
return a || b;
return !a && b;
!a || !b
这就是你所能得到的,尽可能简单的逻辑。当然,您可以选择将变量扩展回其原始定义,但我建议您不要这样做。事实上,我鼓励您不要在
if..return
语句的简单链之外进行简化
如果您使代码更简洁,那么阅读和理解就更具挑战性,这使得调试更具挑战性。很可能我在这篇文章的某个地方“简化”代码时犯了一个错误,在阅读
&&
和|
操作符系列时,如果出现错误,这一点并不明显。当您使用一组链式三元操作符编写代码时,它会变得更加简洁,通常可读性较差
const include =
(options.directory && file !== '.') ? false :
(!dotted) ? true :
(dotted && options.all) ? true :
(dotted && !implied && options.almostall) ? true :
(options.directory && file === '.') ? true :
false;
要对此进行分解,我将首先使用模块模式展开它:
include = (function () {
//set up some simple names for concepts:
var directory = options.directory;
var isDot = file === '.';
var all = options.all;
var almost = options.almostall;
if (directory && !isDot)
return false;
if (!dotted)
return true;
if (dotted && all)
return true;
if (dotted && implied && almost)
return true;
if (directory && isDot)
return true;
return false;
}());
这可以简化。检查后!虚线
,虚线
必须为真,并成为冗余:
转换为:
a
return a;
return a || b;
return !a && b;
!a || !b
作为一件足够好的事情,您可以随意停在这里,知道代码是简单和高效的
当然……这可以简化。最后一个
if
语句可以更改为返回值
:
转换为:
a
return a;
return a || b;
return !a && b;
!a || !b
当然,可以通过将最后一个if
转换为return
来简化:
转换为:
a
return a;
return a || b;
return !a && b;
!a || !b
……而且:
include = (function () {
//set up some simple names for concepts:
var directory = options.directory;
var isDot = file === '.';
var all = options.all;
var almost = options.almostall;
if (directory && !isDot)
return false;
if (!dotted)
return true;
return (all) ||
(implied && almost) ||
(directory && isDot);
}());
include = (function () {
//set up some simple names for concepts:
var directory = options.directory;
var isDot = file === '.';
var all = options.all;
var almost = options.almostall;
if (directory && !isDot)
return false;
return (!dotted) ||
(all) ||
(implied && almost) ||
(directory && isDot);
}());
……而且:
include = (function () {
//set up some simple names for concepts:
var directory = options.directory;
var isDot = file === '.';
var all = options.all;
var almost = options.almostall;
if (directory && !isDot)
return false;
if (!dotted)
return true;
return (all) ||
(implied && almost) ||
(directory && isDot);
}());
include = (function () {
//set up some simple names for concepts:
var directory = options.directory;
var isDot = file === '.';
var all = options.all;
var almost = options.almostall;
if (directory && !isDot)
return false;
return (!dotted) ||
(all) ||
(implied && almost) ||
(directory && isDot);
}());
……而且:
include = (function () {
//set up some simple names for concepts:
var directory = options.directory;
var isDot = file === '.';
var all = options.all;
var almost = options.almostall;
if (directory && !isDot)
return false;
if (!dotted)
return true;
return (all) ||
(implied && almost) ||
(directory && isDot);
}());
include = (function () {
//set up some simple names for concepts:
var directory = options.directory;
var isDot = file === '.';
var all = options.all;
var almost = options.almostall;
if (directory && !isDot)
return false;
return (!dotted) ||
(all) ||
(implied && almost) ||
(directory && isDot);
}());
转换为:
a
return a;
return a || b;
return !a && b;
!a || !b
这可以通过使用以下方法进一步简化:
转换为:
a
return a;
return a || b;
return !a && b;
!a || !b
这就是你所能得到的,尽可能简单的逻辑。当然,您可以选择将变量扩展回其原始定义,但我建议您不要这样做。事实上,我鼓励您不要在
if..return
语句的简单链之外进行简化
如果您使代码更简洁,那么阅读和理解就更具挑战性,这使得调试更具挑战性。很可能我在这篇文章的某个地方“简化”代码时犯了一个错误,在阅读
&&
和|
操作符系列时,如果出现了错误,这一点并不明显。在我看来——尽管有时为复杂的条件提出一行代码看起来确实令人印象深刻,像这样的模式是不可读的,代码也不是很可继承的。我同意优化等等。。但是在我看来,帮助函数中的if..else
在这种情况下保持可读性并没有什么坏处——尽管有时为复杂条件提供一行代码看起来确实令人印象深刻,但像这样的模式是不可读的,代码也不是很可继承的。我同意优化等等。。但在本例中,助手函数中的if..else不会影响保持可读性