1、使用unless简化以下代码到一行
games = ["Super Mario Bros.", "Contra", "Metroid", "Mega Man 2"]
if !games.empty?
puts "Games in your vast collection: #{games.count}"
end
解法：
puts "Games in your vast collection: #{games.count}" unless (games=["Super Mario Bros.", "Contra", "Metroid", "Mega Man 2"]).empty?
2、使用lambda改写以下代码
library = Library.new(GAMES)
print_details = Proc.new do |game|
puts "#{game.name} (#{game.system}) - #{game.year}"
end
library.exec_game('Contra', print_details)
解法：
library = Library.new(GAMES)
print_details = lambda do |game|
puts "#{game.name} (#{game.system}) - #{game.year}"
end
library.exec_game('Contra', print_details)
3、使用yield改写以下代码
def calculation(a, b, &block)
block.call(a, b)
end
puts calculation(5, 5) { |a, b| a + b }
解法：
def calculation(a, b)
yield(a, b)
end
puts calculation(5, 5) { |a, b| a + b }
4、以下代码会输出什么
class C
def hello
@v1 = "var v1 hello"
puts @v1
puts @@v1
end
@v1 = "var v1"
@@v1 = "var v1 C"
puts @v1
end
class D < C
@@v1 = "var v1 D"
end
C.new.hello
C.new.v1
C.v1
解法：
C.new.hello
var v1 hello
var v1 D
C.new.v1
没有定义方法
PS：类方法不能使用实例变量
C.v1
没有定义方法
PS：不能引用一个未初始化的实例变量
5、在使用super关键字的时候 调用 super 和 super() 有什么区别？
解法：
super 当你使用super时，会发送一个消息给当前对象的父类，告诉父类，我的“兄弟”们要调用你的所有参数
super() 不调用参数
6、写一段代码，输出两个时间段时间所有的星期五
start_date = '2017-01-01'
end_date = '2017-05-01'
解法：
start_date = '2017-01-01'
end_date = '2017-05-01'
def puts_Friday
puts
end
7、编写likes方法，实现以下要求，输入字符串数组，输出组合后的字符串
likes [] // must be "no one likes this"
likes ["Peter"] // must be "Peter likes this"
likes ["Jacob", "Alex"] // must be "Jacob and Alex like this"
likes ["Max", "John", "Mark"] // must be "Max, John and Mark like this"
likes ["Alex", "Jacob", "Mark", "Max"] // must be "Alex, Jacob and 2 others like this"
string = "must be"
def likes
puts "no one likes this"
end
8、将一段文字中，每个单词的首字母都大写
输入: "How can mirrors be real if our eyes aren't real"
输出: "How Can Mirrors Be Real If Our Eyes Aren't Real"
解法：
"How can mirrors be real if our eyes aren't real".split.map(&:capitalize).join(' ')
9、写一个方法，判断输入的3个数字能否组成三角形
def isTriangle(a, b, c)
end
解法：
def isTrangle(a, b, c)
if a + b > c && a - b < c
puts "It is a trangle"
end
end
10、写一个方法，只允许使用循环，实现数组反序 [ 1, 2, 3 ] => [ 3, 2, 1 ]
解法：
def bubble_sort
a = [1,2,3]
f = 1
while f < a.length
(0...(a.length-f)).each do |i|
a[i], a[i+1] = a[i+1], a[i] if a[i] < a[i+1]
end
f += 1
end
a
end
PS：