Javascript 变量的三元运算符赋值优先级

问题：当使用

三元

运算符时，它是否计算右操作数的最右项？所以如果我加上逗号，那么它将采用最右边的逗号的表达式

如果我以某种方式反对这个过程，它会评估第二个最右边的术语（Q2）

---

是否所有false子句都分配给变量？如果是这样的话，为什么输出只是

stop.html

，而不包括

Sorry

或

您太年轻了？
我正在写这段代码来演示三元代码

鉴于以下代码：

var url = age > 18 ? (
    alert("OK, you can go."),
    // alert returns "undefined", but it will be ignored because
    // isn't the last comma-separated value of the parenthesis
    "continue.html" // the value to be assigned if age > 18
) : (
    alert("You are much too young!"),
    alert("Sorry :-("), // Q2) ***When does it evaluate this line?
    // etc. etc.
    "stop.html" // the value to be assigned if !(age > 18) Q1) this line is evaluated first right?
);
location.assign(url); // "stop.html"

输出：

// 1. isGreater is called and return true (since true so isLess is never called)
// 2. since true, only times2 is called
var x1 = isGreater(1, 0) || isLess(1, 0) ? times2(5) : times3(5);
console.log('x1 is ' + x1);

// 1. isGreater is called, but return false
// 2. so isLess is called and return true
// 3. since true, only times2 is called
var x2 = isGreater(0, 1) || isLess(0, 1) ? times2(10) : times3(10);
console.log('x2 is ' + x2);

// 1. isGreater is called and return true
// 2. because true, isLess is called and return true
// 3. since both are true, only times2 is called
var x3 = isGreater(1, 0) && isLess(0, 1) ? times2(20) : times3(20);
console.log('x3 is ' + x3);

// 1. isGreater is called, but return false (since false, isLess is never called)
// 2. since false, only times3 is called
var x4 = isGreater(0, 1) && isLess(0, 1) ? times2(30) : times3(30);
console.log('x4 is ' + x4);

或者（| |）场景2

in isGreater
in times2
x1 is 10

in isGreater
in isLess
in times2
x3 is 40

输出：

// 1. isGreater is called and return true (since true so isLess is never called)
// 2. since true, only times2 is called
var x1 = isGreater(1, 0) || isLess(1, 0) ? times2(5) : times3(5);
console.log('x1 is ' + x1);

// 1. isGreater is called, but return false
// 2. so isLess is called and return true
// 3. since true, only times2 is called
var x2 = isGreater(0, 1) || isLess(0, 1) ? times2(10) : times3(10);
console.log('x2 is ' + x2);

// 1. isGreater is called and return true
// 2. because true, isLess is called and return true
// 3. since both are true, only times2 is called
var x3 = isGreater(1, 0) && isLess(0, 1) ? times2(20) : times3(20);
console.log('x3 is ' + x3);

// 1. isGreater is called, but return false (since false, isLess is never called)
// 2. since false, only times3 is called
var x4 = isGreater(0, 1) && isLess(0, 1) ? times2(30) : times3(30);
console.log('x4 is ' + x4);

和（&&）场景1

function isGreater(a, b) {
    console.log('in isGreater');
    return a > b;
}

function isLess(a, b) {
    console.log('in isLess');
    return a < b;
}

function times2(v) {
    console.log('in times2');
    return v * 2;
}

function times3(v) {
    console.log('in times3');
    return v * 3;
}

in isGreater
in isLess
in times2
x2 is 20

输出：

// 1. isGreater is called and return true (since true so isLess is never called)
// 2. since true, only times2 is called
var x1 = isGreater(1, 0) || isLess(1, 0) ? times2(5) : times3(5);
console.log('x1 is ' + x1);

// 1. isGreater is called, but return false
// 2. so isLess is called and return true
// 3. since true, only times2 is called
var x2 = isGreater(0, 1) || isLess(0, 1) ? times2(10) : times3(10);
console.log('x2 is ' + x2);

// 1. isGreater is called and return true
// 2. because true, isLess is called and return true
// 3. since both are true, only times2 is called
var x3 = isGreater(1, 0) && isLess(0, 1) ? times2(20) : times3(20);
console.log('x3 is ' + x3);

// 1. isGreater is called, but return false (since false, isLess is never called)
// 2. since false, only times3 is called
var x4 = isGreater(0, 1) && isLess(0, 1) ? times2(30) : times3(30);
console.log('x4 is ' + x4);

和（&&）场景2

in isGreater
in times2
x1 is 10

in isGreater
in isLess
in times2
x3 is 40

输出：

// 1. isGreater is called and return true (since true so isLess is never called)
// 2. since true, only times2 is called
var x1 = isGreater(1, 0) || isLess(1, 0) ? times2(5) : times3(5);
console.log('x1 is ' + x1);

// 1. isGreater is called, but return false
// 2. so isLess is called and return true
// 3. since true, only times2 is called
var x2 = isGreater(0, 1) || isLess(0, 1) ? times2(10) : times3(10);
console.log('x2 is ' + x2);

// 1. isGreater is called and return true
// 2. because true, isLess is called and return true
// 3. since both are true, only times2 is called
var x3 = isGreater(1, 0) && isLess(0, 1) ? times2(20) : times3(20);
console.log('x3 is ' + x3);

// 1. isGreater is called, but return false (since false, isLess is never called)
// 2. since false, only times3 is called
var x4 = isGreater(0, 1) && isLess(0, 1) ? times2(30) : times3(30);
console.log('x4 is ' + x4);

问：你为什么在这么反常的事情上浪费时间？对于“真实代码”，您希望尽量避免不确定的行为，不是吗？建议：1）将“alert（）”替换为“Console.log（）”，2）在调试器下逐步执行代码，3）确保尝试几种不同的浏览器。例如，Chrome的行为是否与FF/Firebug不同？IE11和IE8不同？你的代码对我来说很好。确保启用了弹出窗口。这只是因为所有语句都已执行，但只分配了最后一次返回。如果要使用
|
运算符而不是
，
则它将在第一个值处停止，并带有一个值。但是是的，这似乎是三元运算符的错误用例。