Why cant I duplicate a slice with copy

Go’s powerful slice type is a fundamental building block for data structures, offering flexibility and efficiency over raw arrays. However, new Go developers often encounter a common pitfall when trying to duplicate a slice with copy(). There’s a widespread misconception that simply using the built-in copy() function on a slice will create a completely independent, new slice with its own distinct underlying data. This isn’t quite how it works, and understanding why is crucial for writing robust and bug-free Go applications. This article will delve into the mechanics of Go slices, clarify the true behavior of the copy() function, and demonstrate the correct approaches to achieve true slice duplication, preventing unexpected side effects in your code.

Understanding Go Slices: More Than Just an Array

To grasp why you can’t truly duplicate a slice with copy() alone, it’s essential to understand what a Go slice fundamentally is. A slice is not an array; rather, it’s a lightweight data structure that provides a view into an underlying array. Think of a slice as a descriptor, or “slice header,” containing three key components: a pointer to the start of the underlying array, its current length (the number of elements it currently holds), and its capacity (the maximum number of elements it can hold without reallocating the underlying array).

When you create a slice, it either references an existing array or a newly created anonymous array. Multiple slices can share the same underlying array, pointing to different segments of it. This design makes slices incredibly efficient for passing around data and creating sub-views without expensive memory allocations. However, this shared underlying array is precisely why a simple copy() operation might not yield the independent duplicate you anticipate.

Consider this: if you have a slice A and create slice B from A using a slicing operation (e.g., B := A[1:3]), both A and B share the same underlying array. Modifying elements in B will directly affect the corresponding elements in A, because they are merely different “windows” into the same data. This shared memory model is at the heart of the confusion surrounding slice duplication.

The copy() Function’s True Purpose

The built-in copy(dst, src []Type) int function in Go is designed for one specific purpose: to copy elements from a source slice to a destination slice. It iterates through the elements of the source slice and places them into the corresponding positions of the destination slice’s underlying array. The number of elements copied is limited by the minimum of the source and destination slice lengths. It returns the number of elements copied.

Crucially, copy() performs a shallow copy of the elements themselves. This means if your slice contains simple types like integers or strings, their values are copied. If your slice contains pointers or struct values, only those pointers or struct values are copied. The data that those pointers might refer to is not recursively copied. This distinction is vital when dealing with complex data structures.

The reason you can’t duplicate a slice with copy() in the sense of creating a completely new, independent slice header with its own separate underlying array is that copy() does not allocate new memory for the destination slice’s underlying array. It expects the destination slice to already exist and have sufficient capacity to receive the copied elements. If the destination slice has insufficient length or capacity, copy() will only copy up to its existing bounds, leaving the impression that it “failed” to duplicate the entire source slice.

To clarify, the copy() function in Go facilitates the transfer of elements between two existing slices. It does not allocate a new underlying array for the destination slice. Instead, it populates the destination slice’s pre-allocated or pre-existing underlying array with values from the source slice, up to the minimum of their respective lengths. This is why a simple call to copy() won’t create a fully independent duplicate slice unless the destination slice was explicitly created with its own new underlying array first.

Why copy() Doesn’t Duplicate the Underlying Array

The core misunderstanding often stems from the expectation that copy() will somehow create a new, separate block of memory for the target slice’s elements. As we’ve established, it simply doesn’t. When you call copy(dst, src), both dst and src are slices, each with their own pointer, length, and capacity. The copy() function uses the pointer of dst to determine where to write the data and the pointer of src to determine where to read the data. It then copies the raw values from src’s underlying array to dst’s underlying array.

Consider a scenario where you have a slice original := []int{1, 2, 3, 4, 5}. If you then declare copyOfSlice := make([]int, 3) and call copy(copyOfSlice, original), only the first three elements (1, 2, 3) will be copied. If you then modify copyOfSlice[0] = 99, the original slice remains unchanged. This is because copyOfSlice was created with make, which allocated a new, distinct underlying array for it. The confusion arises when developers expect copy() to implicitly perform this make operation if the destination slice is not large enough or is a zero-value slice.

The Go language specification is clear about this. As stated in the Go Language Specification, “The copy built-in function copies elements from a source slice into a destination slice. The number of elements copied is the minimum of len(src) and len(dst).” There’s no mention of implicit memory allocation or resizing. This design decision prioritizes explicit memory management and predictable behavior, rather than magical allocations that could lead to performance surprises or obscure bugs. Understanding this distinction is fundamental to effective memory management in Go.

Achieving True Slice Duplication in Go

Since copy() only handles the element transfer, achieving a true, independent duplicate of a slice requires a two-step process: first, ensure the destination slice has its own distinct underlying array of sufficient size, and then use copy() to populate it. Here are the most common and recommended methods:

Method 1: Using make() and copy() (The Standard Approach)

This is the most robust and idiomatic way to create a true duplicate of a slice. You explicitly create a new destination slice with its own underlying array using make(), ensuring it has the same length and capacity as the source slice, and then use copy() to transfer the elements.

Declare the source slice: Define the slice you wish to duplicate.
Create a new destination slice: Use make([]Type, len(sourceSlice), cap(sourceSlice)) to create a new slice. It’s crucial that the length of this new slice matches the length of the source slice if you want to copy all elements. Setting capacity also helps prevent immediate reallocations if the new slice is subsequently extended.
Copy elements: Call copy(destinationSlice, sourceSlice). Since the destination slice now has its own underlying array and sufficient length, all elements from the source slice will be copied, resulting in a fully independent duplicate.

Example:

original := []int{10, 20, 30, 40, 50} duplicate := make([]int, len(original)) // Creates a new underlying array of length 5 copy(duplicate, original) // Now, modifying 'duplicate' will not affect 'original'
<b>Question & Answer : </b><br></br><p>I need to make a copy of a slice in Go and reading the docs there is a <a href="http://golang.org/pkg/builtin/#copy" rel="noreferrer">copy</a> function at my disposal. </p> <blockquote> <p>The copy built-in function copies elements from a source slice into a destination slice. (As a special case, it also will copy bytes from a string to a slice of bytes.) The source and destination may overlap. Copy returns the number of elements copied, which will be the minimum of len(src) and len(dst).</p> </blockquote> <p>But when I do:</p> arr := []int{1, 2, 3} tmp := []int{} copy(tmp, arr) fmt.Println(tmp) fmt.Println(arr)  <p>My tmp is empty as it was before (I even tried to use arr, tmp):</p> [] [1 2 3]  <p>You can check it on go <a href="https://play.golang.org/p/iE6TarF8-S" rel="noreferrer">playground</a>. So why can not I copy a slice?</p>
<br></br><p>The builtin <a href="http://golang.org/pkg/builtin/#copy" rel="noreferrer">copy(dst, src)</a> copies min(len(dst), len(src)) elements.</p> <p>So if your dst is empty (len(dst) == 0), nothing will be copied.</p> <p>Try tmp := make([]int, len(arr)) (<a href="https://play.golang.org/p/xwISeGLzxb" rel="noreferrer">Go Playground</a>):</p> arr := []int{1, 2, 3} tmp := make([]int, len(arr)) copy(tmp, arr) fmt.Println(tmp) fmt.Println(arr)  <p>Output (as expected):</p> [1 2 3] [1 2 3]  <p>Unfortunately this is not documented in the <a href="http://golang.org/pkg/builtin/" rel="noreferrer">builtin</a> package, but it is documented in the <a href="https://golang.org/ref/spec#Appending_and_copying_slices" rel="noreferrer"><strong>Go Language Specification: Appending to and copying slices</strong></a>:</p> <blockquote> <p>The number of elements copied is the minimum of len(src) and len(dst).</p> </blockquote> <p><strong>Edit:</strong></p> <p>Finally the documentation of copy() has been updated and it now contains the fact that the minimum length of source and destination will be copied:</p> <blockquote> <p>Copy returns the number of elements copied, which will be the <strong>minimum</strong> of len(src) and len(dst).</p> </blockquote>