


When does a golang slice operation trigger a new backing array allocation?
Jun 30, 2025 am 01:51 AMIn Go, slicing operations do not necessarily create new underlying arrays. The answer is as follows: 1. When the append operation exceeds the current capacity, a new array will be allocated; 2. Slice the original array within the capacity range; 3. When an independent copy is required, it must be copied explicitly. Specifically, if append causes the length to exceed capacity, the capacity will be expanded as needed during runtime (small slices usually double, and large slices will grow by about 25-50%); multiple slices share the underlying array when accessed within capacity through the s[i:j] syntax; to ensure independent memory, copy() or append techniques are required to force copy, which helps optimize performance and avoid side effects.
In Go, a slice operation doesn't always create a new backing array. It usually reuses the existing one unless certain conditions are met. Knowing when a new allocation happens is important for performance and avoiding unintended side effects.

When appending elements causes capacity to be exceeded
The most common case where a new backing array gets allocated is when you use append()
and there's not enough capacity left in the current slice.

For example:
s := []int{1, 2, 3} s = append(s, 4) // might reuse same array if cap >= 4
If the underlying array has space (ie, len < cap
), the new element fits in the existing array.
But if len == cap
, Go allocates a new array — typically larger than needed — to reduce future allocations.

How much bigger it grows depends on the current size:
- For small slices, it often doubles.
- For large slices, it increases by around 25–50% (exact growth factor may vary between Go versions).
So:
- If you're adding elements and the slice hits its capacity limit → new array is allocated.
- Otherwise, no new allocation happens.
Slicing beyond the current length but within capacity
This is a bit more subtle. You can create a new slice header that points to the same backing array but with a different length or capacity.
Example:
s := []int{1, 2, 3} t := s[1:2] // len=1, cap=2 u := s[1:3] // len=2, cap=2
Both t
and u
share the same underlying array as s
. So modifying an element in one slice affects others.
You can also do this:
s := []int{1, 2, 3} t := s[:4] // allowed if cap >= 4
This doesn't allocate a new array either — it just uses the available capacity.
So:
- As long as you stay within the original array's capacity → no new allocation.
- This is useful for limiting memory usage when working with subsets of data.
Copying a slice explicitly
Sometimes you want to make sure a new backing array is used — for safety or isolation. In this case, you have to do it explicitly , like with copy()
:
src := []int{1, 2, 3} dst := make([]int, len(src)) copy(dst, src)
Now dst
has its own backing array. Modifying src
won't affect dst
.
Alternatively:
dst := append([]int(nil), src...) // creates a new array
Or even:
dst := []int{} copy(dst, src) // works only if dst is already the right size
These approaches let you control when a copy happens. Useful for returning safe results from functions or storing snapshots of mutable data.
Summary
- Appending past capacity triggers new array allocation.
- Slicing within capacity shares the same array.
- To force a new array, use
copy()
orappend()
tricks.
Go gives you flexibility here, but you need to be aware of how slices work under the hood to avoid bugs or optimize performance.
That's basically all you need to know.
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