Ruby unless vs if 逻辑
Ruby unless vs if logic
有人可以解释一下这种行为吗?为什么当 result = true if i
时 nil return 为真,而当 result = false unless i
时 return 为假
unless case screenshot
def my_all?(pattern = nil)
result = true
my_each do |i|
case pattern
when nil
p result, i
result = false unless i
when Regexp
result = false unless i.to_s.match(pattern)
when Class
result = false unless i.is_a?(pattern)
when String, Numeric
result = false unless i == pattern
end
result = yield(i) if block_given? && pattern.nil?
break if !result
end
result
end
if case screenshot
def my_all?(pattern = nil)
result = false
my_each do |i|
case pattern
when nil
p result, i
result = true if i
when Regexp
result = true if i.to_s.match(pattern)
when Class
result = true if i.is_a?(pattern)
when String, Numeric
result = true if i == pattern
end
result = yield(i) if block_given? && pattern.nil?
break if !result
end
result
end
有人可以解释一下这种行为吗?为什么当 result = true if i
时 nil return 为真,而当 result = false unless i
unless case screenshot
def my_all?(pattern = nil)
result = true
my_each do |i|
case pattern
when nil
p result, i
result = false unless i
when Regexp
result = false unless i.to_s.match(pattern)
when Class
result = false unless i.is_a?(pattern)
when String, Numeric
result = false unless i == pattern
end
result = yield(i) if block_given? && pattern.nil?
break if !result
end
result
end
if case screenshot
def my_all?(pattern = nil)
result = false
my_each do |i|
case pattern
when nil
p result, i
result = true if i
when Regexp
result = true if i.to_s.match(pattern)
when Class
result = true if i.is_a?(pattern)
when String, Numeric
result = true if i == pattern
end
result = yield(i) if block_given? && pattern.nil?
break if !result
end
result
end