英文:
do I always have to return a value even on error
问题
如果我有一个看起来像这样的函数：
func ThisIsMyComplexFunc() (ComplexStruct, error)
其中ComplexStruct是一个通常包含大量值的大型结构体。
如果函数在开始之前就遇到错误，还没有开始构建结构体，我希望只返回错误，例如：
return nil, err
但它不允许我这样做，它强制我创建一个虚拟的复杂结构体，并将其与错误一起返回，即使我从不想使用该结构体。
有没有办法解决这个问题？
英文:
If I have a function that looks like this
func ThisIsMyComplexFunc() (ComplexStruct, error)
where ComplexStruct is a big struct that usually contain loads of values.
What if the function stumbles on an error right in the beginning, before I even started building my struct, ideally I would like to only return the error, e.g.
return nil, err
but it wont let me do it, it forces me to create a dummy complex struct and return that together with the error, even though I never want to use the struct.
Is there a way around this?
答案1
得分: 3
如果你的函数声明要返回两个值，那么你需要返回两个值。
一个可能简化你的代码的选项是使用命名返回值：
func ThisIsMyComplexFunc() (s ComplexStruct, err error) {
...
}
现在你可以直接给s和/或err赋值，然后使用一个简单的return语句。你仍然会返回一个ComplexStruct值，但你不需要手动初始化它（它将默认为零值）。
英文:
If your function is declared to return two values, then you will need to return two values.
One option that might simplify your code is to use named return values:
func ThisIsMyComplexFunc() (s ComplexStruct, err error) {
...
}
Now you can just assign to s and/or err and then use a bare return statement. You will still be returning a ComplexStruct value, but you don't need to initialise it manually (it will default to a zero value).
答案2
得分: 2
你可以返回一个指向结构体的指针：
func ThisIsMyComplexFunc() (*ComplexStruct, error) {
...
if somethingIsWrong {
return nil, err
}
...
return &theStructIBuilt, nil
}
一般来说，通过指针传递大的结构体会更加高效，因为复制指针比复制整个结构体更容易。
英文:
You can return a pointer to the struct:
func ThisIsMyComplexFunc() (*ComplexStruct, error) {
...
if somethingIsWrong {
return nil, err
}
...
return &theStructIBuilt, nil
}
In general, it'll be cheaper to pass big structs by pointer anyway, because it's easier to copy a pointer than the whole struct.
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