为什么不能将任意切片赋值给[]interface{}
Introduction
Given that you can assign a variable of any type to an interface{}
, often people will try code like the following.
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var dataSlice []int = foo()
var interfaceSlice []interface{} = dataSlice
This gets the error
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cannot use dataSlice (type []int) as type []interface { } in assignment
The question then, “Why can’t I assign any slice to an []interface{}, when I can assign any type to an interface{}?”
Why?
There are two main reasons for this.
The first is that a variable with type []interface{}
is not an interface! It is a slice whose element type happens to be interface{}
. But even given this, one might say that the meaning is clear.
Well, is it? A variable with type []interface{}
has a specific memory layout, known at compile time.
Each interface{}
takes up two words (one word for the type of what is contained, the other word for either the contained data or a pointer to it). As a consequence, a slice with length N and with type []interface{}
is backed by a chunk of data that is N*2
words long.
This is different than the chunk of data backing a slice with type []MyType
and the same length. Its chunk of data will be N*sizeof(MyType)
words long.
The result is that you cannot quickly assign something of type []MyType
to something of type []interface{}
; the data behind them just look different.
What can I do instead?
It depends on what you wanted to do in the first place.
If you want a container for an arbitrary array type, and you plan on changing back to the original type before doing any indexing operations, you can just use an interface{}
. The code will be generic (if not compile-time type-safe) and fast.
If you really want a []interface{}
because you’ll be doing indexing before converting back, or you are using a particular interface type and you want to use its methods, you will have to make a copy of the slice.
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var dataSlice []int = foo()
var interfaceSlice []interface{} = make([]interface{}, len(dataSlice))
for i, d := range dataSlice {
interfaceSlice[i] = d
}