Is = ["a", "b", "c", "d", "e"] Is.range(1:4:2)为什么错了

根据提供的引用内容，可以看出Is.range(1:4:2)是错误的。这是因为在Java中，没有名为range的方法可以直接应用于数组。如果您想获取数组的子数组，可以使用Arrays类的copyOfRange方法。以下是一个示例代码，展示如何使用copyOfRange方法获取数组的子数组[^1]：

String[] Is = {"a", "b", "c", "d", "e"};
String[] subArray = Arrays.copyOfRange(Is, 1, 4);
System.out.println(Arrays.toString(subArray)); // 输出：[b, c, d]

请注意，copyOfRange方法的第一个参数是源数组，第二个参数是起始索引（包括），第三个参数是结束索引（不包括）。

相关问题

Is = ["a", "b", "c", "d", "e"] Is.range(1:4:2)python中为什么错了

在Python中，列表对象没有`range()`方法。`range()`函数是一个内置函数，用于生成一个整数序列。你可以使用`range()`函数来创建一个整数列表，然后使用切片操作来获取指定范围内的元素。下面是一个示例代码来演示如何使用`range()`函数和切片操作来获取指定范围内的元素[^1]：

Is = ["a", "b", "c", "d", "e"]
result = Is[1:4:2]
print(result)  # 输出：['b', 'd']

在这个例子中，`Is[1:4:2]`表示从索引1开始，到索引4结束（不包括索引4），步长为2。所以结果是包含索引1和索引3的元素，即['b', 'd']。

编写：1. 输入：文法四元组 2. 输出：输出：判定给定文法是否为LL(1)文法

以下是输入文法四元组，判断是否为LL(1)文法的代码实现：

# 输入文法四元组
grammer = {
    'S': ['aA', 'bB'],
    'A': ['cA', 'd'],
    'B': ['eB', 'f']
}
terminals = ['a', 'b', 'c', 'd', 'e', 'f']
non_terminals = ['S', 'A', 'B']

# 构建FIRST集合
first = {}
for symbol in terminals + non_terminals:
    first[symbol] = set()
for terminal in terminals:
    first[terminal].add(terminal)
for non_terminal in non_terminals:
    first[non_terminal] = set()

for non_terminal in non_terminals:
    for production in grammer[non_terminal]:
        if production[0] in terminals:
            first[non_terminal].add(production[0])
        else:
            first[non_terminal] |= first[production[0]]

# 构建FOLLOW集合
follow = {}
for non_terminal in non_terminals:
    follow[non_terminal] = set()
follow['S'].add('$')

while True:
    updated = False
    for non_terminal in non_terminals:
        for production in grammer[non_terminal]:
            for i in range(len(production)):
                if production[i] in non_terminals:
                    if i == len(production) - 1:
                        if follow[non_terminal] != follow[production[i]]:
                            follow[production[i]] |= follow[non_terminal]
                            updated = True
                    else:
                        first_set = first[production[i+1]].copy()
                        if '' in first_set:
                            first_set.remove('')
                            if follow[non_terminal] != follow[production[i]] | first_set:
                                follow[production[i]] |= follow[non_terminal] | first_set
                                updated = True
                        else:
                            if follow[non_terminal] != first_set:
                                follow[production[i]] |= first_set
                                updated = True
    if not updated:
        break

# 构建LL(1)预测分析表
table = {}
for non_terminal in non_terminals:
    table[non_terminal] = {}
    for terminal in terminals:
        table[non_terminal][terminal] = []

for non_terminal in non_terminals:
    for production in grammer[non_terminal]:
        first_set = set()
        for i in range(len(production)):
            if production[i] in terminals:
                first_set = {production[i]}
                break
            else:
                first_set |= first[production[i]]
                if '' not in first[production[i]]:
                    break
        for terminal in first_set:
            table[non_terminal][terminal].append(production)

        if '' in first_set:
            for terminal in follow[non_terminal]:
                table[non_terminal][terminal].append(production)

# 判断文法是否为LL(1)文法
is_ll1_grammer = True
for non_terminal in non_terminals:
    for terminal in terminals:
        if len(table[non_terminal][terminal]) > 1:
            is_ll1_grammer = False
            print(f"({non_terminal}, {terminal}): {table[non_terminal][terminal]}")
print("该文法是LL(1)文法" if is_ll1_grammer else "该文法不是LL(1)文法")

上述代码构建了文法的FIRST集合、FOLLOW集合和LL(1)预测分析表，并通过预测分析表判断该文法是否为LL(1)文法。