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}");
}

我正在

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。

有人可以解释这种行为背后的概念吗？

Answer 1

for 表达式隐式定义了它的迭代变量（这就是你在第二个例子中不需要声明 i 的原因）。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。您的 bind 表达式绑定到 for 循环之外的 curr 变量，而不是 for 中的 curr 变量环。循环外部的那个不会改变，因此绑定的表达式不会改变。

演示 for 的这种行为的示例：

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，如果在 for 循环中修改同名的变量， for 循环外的 curr 将不会改变。< / p>

为什么这个“绑定”代码在JavaFX中没有像预期的那样工作？

1 个答案: