Question

我目前正在使用Busted为lua mod库编写单元测试。有问题的文件定义了一个带有一些函数的模块，然后在底部调用其中一个函数来初始化它自己。

我发现的问题是Busted似乎正在评估所需的文件两次。

测试

it('does a thing', function()
    -- Some setup, replacing globals etc
    require('items')
    assert.are_equal(2, #Items._registry)
end)

模块

Items = { _registry = {} }
function Items.do_some_stuff() end
function some_util_func() end
function load_registry()
  print(debug.traceback())
  for i, itm in pairs(Items.do_some_stuff()) do
    Items._registry[i] = itm
  end
end

load_registry()

正如您所看到的，虽然我已经简化了代码和名称，但结构却并非一帆风顺（据我所知）。

测试将始终失败，因为#Items._registry始终为0（并且转储到控制台会验证该测试）。我尝试在方法内打印，发现它打印了两次;然后我尝试在该功能的顶部使用debug.traceback并找到以下内容。如您所见，堆栈回溯正在打印两次，表明代码正在被评估两次。

这是否是其他任何人遇到的事情？我是否在这种情况下构建错误的测试？或者这是一个错误？

stack traceback:
    items.lua:96: in function 'load_registry'
    items.lua:109: in main chunk
    [C]: in function 'require'
    spec/items_pec.lua:50: in function <spec/items_spec.lua:39>
    [C]: in function 'xpcall'
    /usr/local/share/lua/5.2/busted/core.lua:178: in function 'safe'
    /usr/local/share/lua/5.2/busted/init.lua:40: in function 'executor'
    /usr/local/share/lua/5.2/busted/core.lua:312: in function </usr/local/share/lua/5.2/busted/core.lua:312>
    [C]: in function 'xpcall'
    /usr/local/share/lua/5.2/busted/core.lua:178: in function 'safe'
    ...
    /usr/local/share/lua/5.2/busted/core.lua:312: in function 'execute'
    /usr/local/share/lua/5.2/busted/block.lua:155: in function 'execute'
    /usr/local/share/lua/5.2/busted/init.lua:7: in function 'executor'
    /usr/local/share/lua/5.2/busted/core.lua:312: in function </usr/local/share/lua/5.2/busted/core.lua:312>
    [C]: in function 'xpcall'
    /usr/local/share/lua/5.2/busted/core.lua:178: in function 'safe'
    /usr/local/share/lua/5.2/busted/core.lua:312: in function 'execute'
    /usr/local/share/lua/5.2/busted/execute.lua:58: in function 'execute'
    /usr/local/share/lua/5.2/busted/runner.lua:174: in function </usr/local/share/lua/5.2/busted/runner.lua:11>
    /usr/local/lib/luarocks/rocks/busted/2.0.rc12-1/bin/busted:3: in main chunk
    [C]: in ?
stack traceback:
    items.lua:96: in function 'load_registry'
    items.lua:109: in main chunk
    [C]: in function 'require'
    spec/items_spec.lua:15: in main chunk
    [C]: in function 'xpcall'
    /usr/local/share/lua/5.2/busted/core.lua:178: in function 'safe'
    /usr/local/share/lua/5.2/busted/block.lua:146: in function 'execute'
    /usr/local/share/lua/5.2/busted/init.lua:7: in function 'executor'
    /usr/local/share/lua/5.2/busted/core.lua:312: in function </usr/local/share/lua/5.2/busted/core.lua:312>
    [C]: in function 'xpcall'
    /usr/local/share/lua/5.2/busted/core.lua:178: in function 'safe'
    /usr/local/share/lua/5.2/busted/core.lua:312: in function 'execute'
    /usr/local/share/lua/5.2/busted/execute.lua:58: in function 'execute'
    /usr/local/share/lua/5.2/busted/runner.lua:174: in function </usr/local/share/lua/5.2/busted/runner.lua:11>
    /usr/local/lib/luarocks/rocks/busted/2.0.rc12-1/bin/busted:3: in main chunk
    [C]: in ?

Answer 1

这个问题的答案在于我认为是无关的一些细节但不是（参见我的comment）。

特别是，我将模块的负载行为测试与其各种功能的测试分开。即使在使用busted -t运行以针对特定测试时，也会在两个规范中评估被测模块的导入;即使require调用放在根setup块的describe函数中也是如此。

通过合并两个规格，我能够解决这个双重加载问题。

测试在脚本中立即调用的代码

1 个答案: