为什么这个"绑定"代码在JavaFX中没有像预期的那样工作？

Question

我是JavaFX的新手.我无法理解为什么下面的代码不起作用.

import javafx.util.Sequences;

def nums = [1..10];
var curr = 0;

var evenOrOdd = bind if (nums[curr] mod 2 == 0) "{nums[curr]} is an even number" else "{nums[curr]} is an odd number";

for (curr in [0..(sizeof nums -1)])
{
    println("{evenOrOdd}");
}

我正进入(状态

1 is an odd number
1 is an odd number
1 is an odd number
1 is an odd number
1 is an odd number
1 is an odd number
1 is an odd number
1 is an odd number
1 is an odd number
1 is an odd number

如果我将代码更改为

import javafx.util.Sequences;

def nums = [1..10];
var curr = 0;

var evenOrOdd = bind if (nums[curr] mod 2 == 0) "{nums[curr]} is an even number" else "{nums[curr]} is an odd number";

for (i in [0..(sizeof nums -1)])
{
    curr = i;
    println("{evenOrOdd}");
}

我得到了正确的输出:

1 is an odd number
2 is an even number
3 is an odd number
4 is an even number
5 is an odd number
6 is an even number
7 is an odd number
8 is an even number
9 is an odd number
10 is an even number

显然,循环中的计数器增量不会被视为值更改,并且不会重新计算绑定表达式.

任何人都可以解释这种行为背后的概念吗？

Answer 1

将用于表达含蓄地定义了它的迭代变量(这就是为什么你没有需要声明我在你的第二个例子).即使已经存在具有相同名称的变量,for仍将为其范围创建一个新变量.您绑定表达式绑定到CURR变量之外你的循环,而不是你的内部一个用于循环.循环外部的那个不会改变,因此绑定的表达式不会改变.

例子来说明这种行为的:

var curr = 0;
var ousideCurrRef = bind curr;
println("Before 'for' loop: curr={curr}");
for (curr in [0..3])
{
    println("In 'for' loop: curr={curr} ousideCurrRef={ousideCurrRef}");
}
println("After 'for' loop: curr={curr}");

这将打印:

Before 'for' loop: curr=0
In 'for' loop: curr=0 ousideCurrRef=0
In 'for' loop: curr=1 ousideCurrRef=0
In 'for' loop: curr=2 ousideCurrRef=0
In 'for' loop: curr=3 ousideCurrRef=0
After 'for' loop: curr=0

因此,CURR外的,如果你修改了相同名称的变量里面的循环不会改变的循环.