Escape analysis

Question

In many languages, local variables are located in call stack

In JavaScript/Python, only closure variables are located in heap, because they must live beyond the function calls, they are created.

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In GO, some GO types(like slice type []int) do reference other parts of memory, like JavaScript/Python.

In GO, not all types of variables hold references, like Javascript/Python.

For example,

1) [3]int type variable b directly stores an array of int's, like C, except that C allows to get access of each array element location using C syntax &b[index], for more control

2) int type variable c directly stores an int value, like C, except that, C gives more control by providing syntax(&c) to get location access.

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In GO, my understanding is, for local variables to be on heap/stack depends on applying compiler's escape analysis in example code(below),

func foo() []int {
  // the array lives beyond the call to foo in which it is created
  var a [5]int
  return a[:] // range operator
}

that tells the compiler that variable a lives beyond its scope, so allocate in heap, but not stack.

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Question:

Does the variable a get allocated in heap?

Answer 1

In Go, you are supposed to trust the compiler to make the best decision. It will allocate memory on the stack if possible. See also the FAQ:

From a correctness standpoint, you don't need to know. Each variable in Go exists as long as there are references to it. The storage location chosen by the implementation is irrelevant to the semantics of the language.

The storage location does have an effect on writing efficient programs. When possible, the Go compilers will allocate variables that are local to a function in that function's stack frame. However, if the compiler cannot prove that the variable is not referenced after the function returns, then the compiler must allocate the variable on the garbage-collected heap to avoid dangling pointer errors. Also, if a local variable is very large, it might make more sense to store it on the heap rather than the stack.

In the current compilers, if a variable has its address taken, that variable is a candidate for allocation on the heap. However, a basic escape analysis recognizes some cases when such variables will not live past the return from the function and can reside on the stack.

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Without optimization (inlining), yes a will be allocated in the heap. We can check the escape analysis by passing -gcflags='-m' (https://play.golang.org/p/l3cZFK5QHO):

$ nl -ba 1.go
     1  package main
     2  
     3  func inlined() []int {
     4      var a [5]int
     5      return a[:]
     6  }
     7  
     8  //go:noinline
     9  func no_inline() []int {
    10      var b [5]int
    11      return b[:]
    12  }
    13  
    14  func main() {
    15      var local_array [5]int
    16      var local_var int
    17      println(no_inline())
    18      println(inlined())
    19      println(local_array[:])
    20      println(&local_var)
    21  }

$ go build -gcflags='-m' 1.go
# command-line-arguments
./1.go:3: can inline inlined
./1.go:18: inlining call to inlined
./1.go:5: a escapes to heap
./1.go:4: moved to heap: a
./1.go:11: b escapes to heap
./1.go:10: moved to heap: b
./1.go:18: main a does not escape
./1.go:19: main local_array does not escape
./1.go:20: main &local_var does not escape

We see that the compiler decided to allocate inlined.a on line 5 and no_inline.b on line 10 on the heap, because they both escape their scope.

However, after inlining, the compiler noticed that the a does not escape any more, so it determines the variable can be allocated on the stack again (line 18).

The result is that the variable a is allocated on the main goroutine's stack, while the variable b is allocated on the heap. As we can see from the output, the address of b is on 0x1043xxxx while all other are on 0x1042xxxx.

$ ./1
[5/5]0x10432020
[5/5]0x10429f58
[5/5]0x10429f44
0x10429f40