javascript - 三元运算符对变量的赋值优先级

Question

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"

问题:当使用三元 运算符时，它是否计算右操作数的最右项？所以如果我输入逗号，那么它会采用最右边逗号的表达式。

如果我以某种方式调用此过程，它会评估最右边的第二个项 (Q2)。

是不是所有的false子句都赋值给了变量？如果是这样，为什么输出只有 stop.html 而没有包含 Sorry 或 You are much too young!？

Answer 1

我正在编写这个小代码块来演示现在的三元运算。

给定以下代码:

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;
}

或 (||) 场景 1

// 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);

输出:

in isGreater
in times2
x1 is 10

或 (||) 场景 2

// 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);

输出:

in isGreater
in isLess
in times2
x2 is 20

和(&&)场景1

// 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);

输出:

in isGreater
in isLess
in times2
x3 is 40

和(&&)场景2

// 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);

输出:

in isGreater
in times3
x4 is 90