9 ms·

One of the things I think the Go tutorials don't make a big enough deal of is that Go is relatively explicit about allocations. := isn't just a shortcut for dec

by jerf 6y ago

One of the things I think the Go tutorials don't make a big enough deal of is that Go is relatively explicit about allocations. := isn't just a shortcut for declaring variables, it's an allocation, and an error to use it when it doesn't allocate. var X Sometype isn't just a declaration, it's an allocation. := kinda smears the clarity by not allocating if you have a variable on the left that is already allocated, and there's some other places where it kinda smears things up, but at the core, Go makes you explicitly allocate.

gowld 6y ago
I think the way to say it is that Go requires you to declare every allocation, but allows over-declaration in the case of copying. > := [...] an error to use it when it doesn't allocate. > := [...] not allocating if you have a variable on the left that is already allocated, This appears to be a contradiction. I suppose you mean something like "error to use it when there's no possible context where that line of code would allocate"; what's an example of that?
- fred256 6y ago
  a, b := 1, 2 If either a or b (but not both) were already defined, this won't re-define (and reallocate space for) them.
  - gowld 6y ago
    Aha, lossy compression syntax!
catlifeonmars 6y ago
I’m not sure I understood this correctly. Does the following allocate (on the heap)? foo := MyStruct{}
- kevincox 6y ago
  No. In this case foo will live on the stack (unless you take its address later). foo := MyStruct{} // Could live on the stack _ := &foo // Oh, now foo must live on the heap.
  - coder543 6y ago
    This isn't even true. No matter what syntax you write inside a function, the Go compiler always has the final say on what is stack allocated and what is heap allocated. Taking the address of foo will not cause foo to be heap allocated unless Go is unable to prove that the pointer will live for less time than the current stack frame. Look up "escape analysis". Basically the only way to guarantee that something will always be heap allocated is to assign it to a global variable. Even returning a pointer to that object from the current function is not a strong guarantee, since the compiler could inline this function into the caller and determine that everything can live happily inside the newly inlined stack frame without heap allocation.
    - kevincox 6y ago
      Good point. It isn't "must", I should have said "may".
- coder543 6y ago
  No, that does not cause a heap allocation on its own. If other lines of code in that function caused a pointer to that value to escape the lifetime of the current function's stack frame, the compiler would determine that it has to be heap allocated instead. I believe the person you are replying to was making a confusing point about some hand wavy notion of "any kind of allocation", which includes stack allocations... which are determined at compile time, not with "alloca".
  - catlifeonmars 6y ago
    Ah, that makes much more sense!
kevincox 6y ago
I don't see how: temp := someFunc() p := &temp is any less explicit than p := &someFunc() It seems that the `&` is still required to put something onto the heap.
- zeeboo 6y ago
  What about `&m[x]` where m is some map? Does that heap allocate and create a copy, or is it a pointer to the actual storage slot? If the former, that's a hidden copy/allocation that didn't exist before, and if it's the latter, resizing the map invalidates the pointer, so it must be updated somehow.
  - kevincox 6y ago
    `&` will "move" something to the heap if it isn't already on the heap. The simpler way to think about it is that in Golang everything is on the heap. However the optimizer will move things to the stack if they don't have their address taken. I think the point about explicitness is that if you don't use `&` then it will be able to be put on the stack. So `&` doesn't cause a heap allocation but lack of `&` (or new()) confirms that there isn't one. (I don't actually know if that is true but I can't think of any counterexamples)
    - zeeboo 6y ago
      I think I didn't communicate my point clearly. Consider this hypothetical program: x := make(map[int]int) x[0] = 5 y := &x[0] *y = 10 print(x[0]) // 5 or 10? x[0] = 6 print(*y) // 6 or 10? // force the map to grow and reallocate the buckets for j := 1; j < 100; j++ { x[j] = j } *y = 11 print(x[0]) // 5, 6, 10, or 11? The crux of the problem is answering what y actually points at: the value in the map bucket, or some freshly allocated value? There are problems with whichever one you pick. edit: changed the second print to *y instead of x[0]. thanks masklinn for catching this error.
      - masklinn 6y ago
coder543 6y ago
> it's an allocation Most people probably think "heap allocation" when you say this. Go doesn't do dynamic allocations within a stack frame (alloca in C), so when you say "it's an allocation", what does that mean? It could be a stack allocation that occurred at compile time as a reservation in the size of the stack frame for that function. It could be a heap allocation. Only the compiler knows! The Go compiler is the ultimate authority on what becomes a heap allocation. It tries to make everything into a stack allocation when possible, and stack allocations are "free". Beyond that, a sufficiently smart compiler can reuse stack "allocations" within a single function as certain values become "dead" (never used again). So there isn't even guaranteed to be a 1:1 correspondence between "stack allocations" and variables that you declared inside the function. So, I completely disagree with your statement about Go being "relatively explicit about allocations." It's one of the least explicit compiled languages in that regard. Go makes a distinction between declaration and assignment, which is the syntax you're talking about. It really has nothing to do with allocations.
- jerf 6y ago
  You file a good complaint, and I should clarify. What I mean is more like Go is secretly quite explicit about its allocations, if I may. It superficially looks like it doesn't really care with a variety of syntax glosses that can make it look like it's more like a scripting language where it doesn't care, but it actually does care quite a lot even at the syntax level, and if you dig past the syntax glosses, it is actually explicit about what gets allocated. It doesn't successfully hide it from you like a scripting language does. Also, allocations are just... allocations. Go qua Go doesn't have stack vs. heap, and it's a mistake to care except when optimizing. So in Go qua Go, it isn't an issue that it may "reuse" a particular address, because in Go qua Go you can't witness that anyhow. (If you try to keep a pointer around to witness it with, you'll keep the thing pointed to alive.) From Go's perspective, it's still an allocation even if the implementation manages to re-use a particular memory address to do so. I'm talking about the runtime Go implements here, not the implementation. This actually took me some years to correctly internalize, for what it's worth. It does a "good" job of glossing over things. However, if you really poke at it, allocations are still explicit. They just may not look like what you are used to from other languages.

by jerf 6y ago

gowld 6y ago

I think the way to say it is that Go requires you to declare every allocation, but allows over-declaration in the case of copying. > := [...] an error to use it when it doesn't allocate. > := [...] not allocating if you have a variable on the left that is already allocated, This appears to be a contradiction. I suppose you mean something like "error to use it when there's no possible context where that line of code would allocate"; what's an example of that?

fred256 6y ago
a, b := 1, 2 If either a or b (but not both) were already defined, this won't re-define (and reallocate space for) them.
- gowld 6y ago
  Aha, lossy compression syntax!

catlifeonmars 6y ago

I’m not sure I understood this correctly. Does the following allocate (on the heap)? foo := MyStruct{}

kevincox 6y ago
No. In this case foo will live on the stack (unless you take its address later). foo := MyStruct{} // Could live on the stack _ := &foo // Oh, now foo must live on the heap.
- coder543 6y ago
  This isn't even true. No matter what syntax you write inside a function, the Go compiler always has the final say on what is stack allocated and what is heap allocated. Taking the address of foo will not cause foo to be heap allocated unless Go is unable to prove that the pointer will live for less time than the current stack frame. Look up "escape analysis". Basically the only way to guarantee that something will always be heap allocated is to assign it to a global variable. Even returning a pointer to that object from the current function is not a strong guarantee, since the compiler could inline this function into the caller and determine that everything can live happily inside the newly inlined stack frame without heap allocation.
  - kevincox 6y ago
    Good point. It isn't "must", I should have said "may".
coder543 6y ago
No, that does not cause a heap allocation on its own. If other lines of code in that function caused a pointer to that value to escape the lifetime of the current function's stack frame, the compiler would determine that it has to be heap allocated instead. I believe the person you are replying to was making a confusing point about some hand wavy notion of "any kind of allocation", which includes stack allocations... which are determined at compile time, not with "alloca".
- catlifeonmars 6y ago
  Ah, that makes much more sense!

kevincox 6y ago

I don't see how: temp := someFunc() p := &temp is any less explicit than p := &someFunc() It seems that the `&` is still required to put something onto the heap.

zeeboo 6y ago
What about `&m[x]` where m is some map? Does that heap allocate and create a copy, or is it a pointer to the actual storage slot? If the former, that's a hidden copy/allocation that didn't exist before, and if it's the latter, resizing the map invalidates the pointer, so it must be updated somehow.
- kevincox 6y ago
  `&` will "move" something to the heap if it isn't already on the heap. The simpler way to think about it is that in Golang everything is on the heap. However the optimizer will move things to the stack if they don't have their address taken. I think the point about explicitness is that if you don't use `&` then it will be able to be put on the stack. So `&` doesn't cause a heap allocation but lack of `&` (or new()) confirms that there isn't one. (I don't actually know if that is true but I can't think of any counterexamples)
  - zeeboo 6y ago
    I think I didn't communicate my point clearly. Consider this hypothetical program: x := make(map[int]int) x[0] = 5 y := &x[0] *y = 10 print(x[0]) // 5 or 10? x[0] = 6 print(*y) // 6 or 10? // force the map to grow and reallocate the buckets for j := 1; j < 100; j++ { x[j] = j } *y = 11 print(x[0]) // 5, 6, 10, or 11? The crux of the problem is answering what y actually points at: the value in the map bucket, or some freshly allocated value? There are problems with whichever one you pick. edit: changed the second print to *y instead of x[0]. thanks masklinn for catching this error.
    - masklinn 6y ago

coder543 6y ago

> it's an allocation Most people probably think "heap allocation" when you say this. Go doesn't do dynamic allocations within a stack frame (alloca in C), so when you say "it's an allocation", what does that mean? It could be a stack allocation that occurred at compile time as a reservation in the size of the stack frame for that function. It could be a heap allocation. Only the compiler knows! The Go compiler is the ultimate authority on what becomes a heap allocation. It tries to make everything into a stack allocation when possible, and stack allocations are "free". Beyond that, a sufficiently smart compiler can reuse stack "allocations" within a single function as certain values become "dead" (never used again). So there isn't even guaranteed to be a 1:1 correspondence between "stack allocations" and variables that you declared inside the function. So, I completely disagree with your statement about Go being "relatively explicit about allocations." It's one of the least explicit compiled languages in that regard. Go makes a distinction between declaration and assignment, which is the syntax you're talking about. It really has nothing to do with allocations.

jerf 6y ago
You file a good complaint, and I should clarify. What I mean is more like Go is secretly quite explicit about its allocations, if I may. It superficially looks like it doesn't really care with a variety of syntax glosses that can make it look like it's more like a scripting language where it doesn't care, but it actually does care quite a lot even at the syntax level, and if you dig past the syntax glosses, it is actually explicit about what gets allocated. It doesn't successfully hide it from you like a scripting language does. Also, allocations are just... allocations. Go qua Go doesn't have stack vs. heap, and it's a mistake to care except when optimizing. So in Go qua Go, it isn't an issue that it may "reuse" a particular address, because in Go qua Go you can't witness that anyhow. (If you try to keep a pointer around to witness it with, you'll keep the thing pointed to alive.) From Go's perspective, it's still an allocation even if the implementation manages to re-use a particular memory address to do so. I'm talking about the runtime Go implements here, not the implementation. This actually took me some years to correctly internalize, for what it's worth. It does a "good" job of glossing over things. However, if you really poke at it, allocations are still explicit. They just may not look like what you are used to from other languages.