Question

这是一个最小的工作示例：

public class FF {

    @Test
    public void test01() {
    final ListProperty p = new SimpleListProperty(FXCollections.observableArrayList());
    p.addListener((ListChangeListener) c -> {
        System.err.println("Listener here..");
    });
    Bindings.createObjectBinding(() -> {
        System.err.println("Binding here");
        return null;
    }, p);

    p.add("hans");
    }

    @Test
    public void test02() {
    final ListProperty p = new SimpleListProperty(FXCollections.observableArrayList());
    final ListProperty p2 = new SimpleListProperty(FXCollections.observableArrayList());
    p.addListener((ListChangeListener) c -> {
        System.err.println("Listener 1 here..");
    });
    p2.addListener((ListChangeListener) c -> {
        System.err.println("Listener 2 here..");
    });
    final ObjectBinding ob = Bindings.createObjectBinding(() -> {
        System.err.println("Binding here");
        return null;
    }, p);
    p2.bind(ob);
    p.add("hans");

    }

}

第二次测试看起来像预期，但对于第一次测试，输出只是“Listener here ..”。为什么绑定在这种情况下不起作用？

匿名侦听器和匿名绑定有什么区别？

Answer 1

你已经创建了Binding对象但没有通过它绑定任何东西，因此它优化了自己什么都不做。

添加任意随机操作以查看“此处绑定...”输出。 E.g：

        ###########################################################
        #  Algorithm
        #    Ask for a word
        #     Count and store all vowels in the words (repeats only are 
        #      counted once)
        #     count and store only consonants after the last vowel, if the 
        #     if last letter is a vowel no consonants are stored
        #       once all 5 vowels are stored or at least 5 consonants are 
        #       stored 
        #           print
        #               what vowels appear and how many 
        #               what consonants appear   
        ###########################################################

VOWELS = 'aeiou'

word = input('Input a word: ')
wordlow = word.lower() #converts the input to all lowercase

vowcount = 0
concount = 0

vowcollected_str = ''
concollected_str = ''

#ends the program once 5 vowels or consonants have been stored
while vowcount <= 4 and concount <= 4: 
    vowcount = 0
    concount = 0

    #stores the actual letter that is being stored, not just how many
    vowcollected_str = ''
    concollected_str = ''

    for i, ch in enumerate(wordlow):
        if ch in VOWELS:
            if ch not in vowcollected_str:
                vowcollected_str += ch 
                position = i
                vowcount += len(vowcollected_str)
        if ch not in VOWELS:
                if ch not in concollected_str:
                    concollected_str += wordlow[i:]
                    concount += len(word[i:]) 
    word = input('Input a word: ')
    wordlow = word.lower()
print(vowcount)
print(vowcollected_str) 
print(concount)
print(concollected_str)

听众工作，绑定没有

1 个答案: