Allocating on the Stack

Hacker News · Feb 27, 2026 · Collected from RSS

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We’re always looking for ways to make Go programs faster. In the last 2 releases, we have concentrated on mitigating a particular source of slowness, heap allocations. Each time a Go program allocates memory from the heap, there’s a fairly large chunk of code that needs to run to satisfy that allocation. In addition, heap allocations present additional load on the garbage collector. Even with recent enhancements like Green Tea, the garbage collector still incurs substantial overhead. So we’ve been working on ways to do more allocations on the stack instead of the heap. Stack allocations are considerably cheaper to perform (sometimes completely free). Moreover, they present no load to the garbage collector, as stack allocations can be collected automatically together with the stack frame itself. Stack allocations also enable prompt reuse, which is very cache friendly. Stack allocation of constant-sized slices Consider the task of building a slice of tasks to process: func process(c chan task) { var tasks []task for t := range c { tasks = append(tasks, t) } processAll(tasks) } Let’s walk through what happens at runtime when pulling tasks from the channel c and adding them to the slice tasks. On the first loop iteration, there is no backing store for tasks, so append has to allocate one. Because it doesn’t know how big the slice will eventually be, it can’t be too aggressive. Currently, it allocates a backing store of size 1. On the second loop iteration, the backing store now exists, but it is full. append again has to allocate a new backing store, this time of size 2. The old backing store of size 1 is now garbage. On the third loop iteration, the backing store of size 2 is full. append again has to allocate a new backing store, this time of size 4. The old backing store of size 2 is now garbage. On the fourth loop iteration, the backing store of size 4 has only 3 items in it. append can just place the item in the existing backing store and bump up the slice length. Yay! No call to the allocator for this iteration. On the fifth loop iteration, the backing store of size 4 is full, and append again has to allocate a new backing store, this time of size 8. And so on. We generally double the size of the allocation each time it fills up, so we can eventually append most new tasks to the slice without allocation. But there is a fair amount of overhead in the “startup” phase when the slice is small. During this startup phase we spend a lot of time in the allocator, and produce a bunch of garbage, which seems pretty wasteful. And it may be that in your program, the slice never really gets large. This startup phase may be all you ever encounter. If this code was a really hot part of your program, you might be tempted to start the slice out at a larger size, to avoid all of these allocations. func process2(c chan task) { tasks := make([]task, 0, 10) // probably at most 10 tasks for t := range c { tasks = append(tasks, t) } processAll(tasks) } This is a reasonable optimization to do. It is never incorrect; your program still runs correctly. If the guess is too small, you get allocations from append as before. If the guess is too large, you waste some memory. If your guess for the number of tasks was a good one, then there’s only one allocation site in this program. The make call allocates a slice backing store of the correct size, and append never has to do any reallocation. The surprising thing is that if you benchmark this code with 10 elements in the channel, you’ll see that you didn’t reduce the number of allocations to 1, you reduced the number of allocations to 0! The reason is that the compiler decided to allocate the backing store on the stack. Because it knows what size it needs to be (10 times the size of a task) it can allocate storage for it in the stack frame of process2 instead of on the heap1. Note that this depends on the fact that the backing store does not escape to the heap inside of processAll. Stack allocation of variable-sized slices But of course, hard coding a size guess is a bit rigid. Maybe we can pass in an estimated length? func process3(c chan task, lengthGuess int) { tasks := make([]task, 0, lengthGuess) for t := range c { tasks = append(tasks, t) } processAll(tasks) } This lets the caller pick a good size for the tasks slice, which may vary depending on where this code is being called from. Unfortunately, in Go 1.24 the non-constant size of the backing store means the compiler can no longer allocate the backing store on the stack. It will end up on the heap, converting our 0-allocation code to 1-allocation code. Still better than having append do all the intermediate allocations, but unfortunate. But never fear, Go 1.25 is here! Imagine you decide to do the following, to get the stack allocation only in cases where the guess is small: func process4(c chan task, lengthGuess int) { var tasks []task if lengthGuess <= 10 { tasks = make([]task, 0, 10) } else { tasks = make([]task, 0, lengthGuess) } for t := range c { tasks = append(tasks, t) } processAll(tasks) } Kind of ugly, but it would work. When the guess is small, you use a constant size make and thus a stack-allocated backing store, and when the guess is larger you use a variable size make and allocate the backing store from the heap. But in Go 1.25, you don’t need to head down this ugly road. The Go 1.25 compiler does this transformation for you! For certain slice allocation locations, the compiler automatically allocates a small (currently 32-byte) slice backing store, and uses that backing store for the result of the make if the size requested is small enough. Otherwise, it uses a heap allocation as normal. In Go 1.25, process3 performs zero heap allocations, if lengthGuess is small enough that a slice of that length fits into 32 bytes. (And of course that lengthGuess is a correct guess for how many items are in c.) We’re always improving the performance of Go, so upgrade to the latest Go release and be surprised by how much faster and memory efficient your program becomes! Stack allocation of append-allocated slices Ok, but you still don’t want to have to change your API to add this weird length guess. Anything else you could do? Upgrade to Go 1.26! func process(c chan task) { var tasks []task for t := range c { tasks = append(tasks, t) } processAll(tasks) } In Go 1.26, we allocate the same kind of small, speculative backing store on the stack, but now we can use it directly at the append site. On the first loop iteration, there is no backing store for tasks, so append uses a small, stack-allocated backing store as the first allocation. If, for instance, we can fit 4 tasks in that backing store, the first append allocates a backing store of length 4 from the stack. The next 3 loop iterations append directly to the stack backing store, requiring no allocation. On the 4th iteration, the stack backing store is finally full and we have to go to the heap for more backing store. But we have avoided almost all of the startup overhead described earlier in this article. No heap allocations of size, 1, 2, and 4, and none of the garbage that they eventually become. If your slices are small, maybe you will never have a heap allocation. Stack allocation of append-allocated escaping slices Ok, this is all good when the tasks slice doesn’t escape. But what if I’m returning the slice? Then it can’t be allocated on the stack, right? Right! The backing store for the slice returned by extract below can’t be allocated on the stack, because the stack frame for extract disappears when extract returns. func extract(c chan task) []task { var tasks []task for t := range c { tasks = append(tasks, t) } return tasks } But you might think, the returned slice can’t be allocated on the stack. But what about all those intermediate slices that just become garbage? Maybe we can allocate those on the stack? func extract2(c chan task) []task { var tasks []task for t := range c { tasks = append(tasks, t) } tasks2 := make([]task, len(tasks)) copy(tasks2, tasks) return tasks2 } Then the tasks slice never escapes extract2. It can benefit from all of the optimizations described above. Then at the very end of extract2, when we know the final size of the slice, we do one heap allocation of the required size, copy our tasks into it, and return the copy. But do you really want to write all that additional code? It seems error prone. Maybe the compiler can do this transformation for us? In Go 1.26, it can! For escaping slices, the compiler will transform the original extract code to something like this: func extract3(c chan task) []task { var tasks []task for t := range c { tasks = append(tasks, t) } tasks = runtime.move2heap(tasks) return tasks } runtime.move2heap is a special compiler+runtime function that is the identity function for slices that are already allocated in the heap. For slices that are on the stack, it allocates a new slice on the heap, copies the stack-allocated slice to the heap copy, and returns the heap copy. This ensures that for our original extract code, if the number of items fits in our small stack-allocated buffer, we perform exactly 1 allocation of exactly the right size. If the number of items exceeds the capacity our small stack-allocated buffer, we do our normal doubling-allocation once the stack-allocated buffer overflows. The optimization that Go 1.26 does is actually better than the hand-optimized code, because it does not require the extra allocation+copy that the hand-optimized code always does at the end. It requires the allocation+copy only in the case that we’ve exclusively operated on a stack-backed slice up to the return point. We do pay the cost for a copy, but that cost is almost completely offset by the copies in the startup phase that we no longer have to do. (In fact, the the new scheme at worst has to copy one more element than the old scheme.) Wrapping up Hand optimization can still be beneficial, especially if you have a good estim