Question

以下函数getTank()有效（如果不用作打印fluidName的函数）。当我返回值然后尝试在函数外部访问它们时会出现问题。结果是'attempt to concatenate string and nil'在行mon2.write(returnedVariable)处，例如在函数之外。

如果我只是执行以下操作：

for k,v in pairs(tableInfo) do amount=v.amount end 
print(amount)

在函数之外，它给出了正确的值。

function getTank(tankPeriph)
    -- This has been tested and works

    local tableInfo = tankPeriph.getTankInfo("unknown")
    local fluidRaw, fluidName, fluidAmount, fluidCapacity

    for k,v in pairs(tableInfo) do
        fluidRaw = v.rawName
        fluidName = v.name
        fluidAmount = v.amount
        fluidCapacity = v.capacity
    end
    return fluidRaw, fluidName, fluidAmount, fluidCapacity
end

function dispTanks()
    -- working on it

    -- TANK 0
    mon2.setCursorPos(rowPos, ironTank0Col)
    mon2.clearLine()
    local fluidRaw, fluidName, fluidAmount, fluidCapacity = getTank(irontank0)
    mon2.write("Iron Tank 0 (" .. fluidName .. ") : " .. fluidAmount)

    -- TANK 1
    mon2.setCursorPos(rowPos, ironTank1Col)
    mon2.clearLine()
    local fluidRaw, fluidName, fluidAmount, fluidCapacity = getTank(irontank1)
    mon2.write("Iron Tank 1 (" .. fluidName .. ") : " .. fluidAmount)

    -- TANK 2
    mon2.setCursorPos(rowPos, ironTank2Col)
    mon2.clearLine()
    local fluidRaw, fluidName, fluidAmount, fluidCapacity = getTank(irontank2)
    mon2.write("Iron Tank 2 (" .. fluidName .. ") : " .. fluidAmount)

    -- TANK 3
    mon2.setCursorPos(rowPos, ironTank3Col)
    mon2.clearLine()
    local fluidRaw, fluidName, fluidAmount, fluidCapacity = getTank(irontank3)
    mon2.write("Iron Tank 3 (" .. fluidName .. ") : " .. fluidAmount)

    -- TANK 4
    mon2.setCursorPos(rowPos, ironTank4Col)
    mon2.clearLine()
    local fluidRaw, fluidName, fluidAmount, fluidCapacity = getTank(irontank4)
    mon2.write("Iron Tank 4 (" .. fluidName .. ") : " .. fluidAmount)

end

Answer 1

function getTank(tankPeriph)
    -- This has been tested and works

    local tableInfo = tankPeriph.getTankInfo("unknown") -- Local to the getTank function.

    for k,v in pairs(tableInfo) do
        local fluidRaw = v.rawName -- local to this for loop
        local fluidName = v.name -- local to this for loop
        local fluidAmount = v.amount -- local to this for loop
        local fluidCapacity = v.capacity -- local to this for loop
    end

    return fluidRaw, fluidName, fluidAmount, fluidCapacity -- Returning the values of global variables (which are nil).
end

正如我上面编辑的代码段中所示，您的本地人不在您认为的位置本地，并且您没有正确地从您的函数返回其值。将这些变量的局部声明移到for循环之外（如果需要的话，将赋值保留在for循环中，尽管我无法想象你这样做，因为你只得到循环中的最后一个值）和你的功能应“工作”。

Answer 2

＆＃34; local＆＃34;限定符将范围限制为阻塞或块，因此getTank（）循环中的locals限定为循环;在循环外，他们的价值观会丢失。因此，当getTank返回时，它返回的变量尚未在函数范围内定义，因此它们都是零。有关有用示例，请参阅http://www.lua.org/manual/5.1/manual.html#2.6。

但是，由于这似乎无法解决您的问题，我打赌您还有一个额外的问题，即local tableInfo是空表，这意味着tankPeriph.getTankInfo("unknown")返回空表（不是{{ 1}}，但是nil）。

Lua函数多个返回值，打印（returnvalue）导致尝试连接字符串和Nil

2 个答案: