如何更改 lua 中的原始变量

Question

我想更改原始变量，以便函数打印不同的答案。我是 Lua 的新手，这可能吗？我该怎么做？

这是我的代码：

io.write("Hello, what is your first player's name? ")
local name = io.read()
io.write('How old is Player 1? ')
local age = io.read()

function Recall()
    print("Your player's name: " ,name, "\nYour player's age: " ,age, "\n")
end

Recall()

io.write("What is your second player's name? ")
local name = io.read()
io.write('How old is Player 2? ')
local age = io.read()

Recall()

当我第二次调用该函数时，它会在第一次输入时打印姓名和年龄。有什么建议吗？

Answer 1

您创建了新的本地人并失去了对同名本地人的访问权限。

要解决此问题，请移除第二个 name 和第二个 age 附近的 local。（在 Player2 附近）。

作为替代解决方案，您可以为 Recall 函数设置参数并在其中传递参数。

Answer 2

好主意，稍微调整一下，这样您就不会使用相同的变量

io.write("Hello, what is your first player's name? ")
local name1 = io.read()
io.write('How old is Player 1? ')
local age1 = io.read()

function Recall( name, age )
    print("Your player's name: ", name, "\nYour player's age: ", age, "\n")
end

Recall( name1, age1 )

io.write("What is your second player's name? ")
local name2 = io.read()
io.write('How old is Player 2? ')
local age2 = io.read()

Recall( name2, age2 )

如何更改 lua 中的原始变量

How can I change the original variable in lua

variables

lua

var

function