三元运算符赋值优先级为变量

Ternary Operator Assignment Precedence to Variable

本文关键字：赋值 优先级 变量 运算符 三元      更新时间：2023-09-26
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：使用 ternary 运算符时，它是否计算正确操作数的最右边项？ 因此，如果我输入逗号，那么它将采用最右边逗号的表达式。

如果我以某种方式呼吁这个过程反对，它会评估第二个最右边的术语(Q2(。

是否将所有 false 子句都分配给变量？如果是这样，为什么输出只有stop.html，不包括Sorry或You are much too young!？

我正在编写这一小段代码来演示现在的三元作品。

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
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