lua 语句学习

就如同C里的if else,while,do,repeat。就看lua里怎么用：

1、首先看if else

t = {1,2,3}

local i = 1

if t[i] and t[i] % 2 == 0 then
	print("even")
else
	print("odd")
end

2、while

while t[i] do
	print(t[i])
	i = i + 1
end

do
	print(t[i])
	i = i + 1
end

local i = 1
repeat
	print(t[i])
	i = i + 1
until t[i] == nil

for i = 1, #t do
	print(t[i])
end

for k,v in pairs(t) do
	print(k , v)
end

for k,v in pairs(t) do
	print(k , v)
	if k == 2 then
	break
	end
end

1、lua为毛支持elseif这样的写法

比方能够这么写：

for k,v in pairs(t) do
	if k == 2 then
		print(‘111‘)
	elseif k == 1 then
		print(k , v)
	end
end

t = {1,2,3}


for k,v in pairs(t) do
	if k == 2 then
		print(‘111‘)
	else if k == 1 then
		print(k , v)
		end
	end
end

由于lua里没有switch，假设选择分支比較多写那么多end非常丑陋，so就能够elseif一起用啦。

2、写无条件运行代码，假设c里我喜欢用do{}while(false)；lua就能够有非常多写法，来个最easy想到的：

local i = 0
repeat
	i = i + 1
	print(i)
	if i > 10 then
		break
	end
until false

4、改动恶心的goto语句，的确看着非常不爽。改了就清爽多了：

function room1()
repeat
	local move = io.read()
	if move == "south" then
	return room3()
	elseif move == "east" then
	return room2()
	else
	print("invalid move")
	end
until false
end

function room2()
repeat
	local move = io.read()
	if move == "south" then
	return room4()
	elseif move == "west" then
	return room1()
	else
	print("invalid move")
	end
until false
end

function room3()
repeat
	local move = io.read()
	if move == "north" then
	return room1()
	elseif move == "east" then
	return room4()
	else
	print("invalid move")
	end
until false
end

function room4()
	print("Congratulations, u win!")
end

room1()

goto不能调到某个语句块内。由于内部是不能为外部所知道的。为毛不能跳出函数块，书里的解释是：

stackflow的答案是：

Your guesses are hinting at the answer. The reason is because the?goto?statement and its destination must reside in the same stack frame. The program context before and after the?goto?need to be the same otherwise the code being jumped to won‘t be running in its correct stack frame and its behavior will be undefined.?goto?in C has the same restrictions for the same reasons.

6、这道题看到goto就好凌乱，我压根就打算用goto。也驾驭不了介个，so不分析了。