Golang append memory allocation VS. STL push_back memory allocation - memory

Question

I compared the Go append function and the STL vector.push_back and found that different memory allocation strategy which confused me. The code is as follow:
// CPP STL code
void getAlloc() {
vector<double> arr;
int s = 9999999;
int precap = arr.capacity();
for (int i=0; i<s; i++) {
if (precap < i) {
arr.push_back(rand() % 12580 * 1.0);
precap = arr.capacity();
printf("%d %p\n", precap, &arr[0]);
} else {
arr.push_back(rand() % 12580 * 1.0);
}
}
printf("\n");
return;
}
// Golang code
func getAlloc() {
arr := []float64{}
size := 9999999
pre := cap(arr)
for i:=0; i<size; i++ {
if pre < i {
arr = append(arr, rand.NormFloat64())
pre = cap(arr)
log.Printf("%d %p\n", pre, &arr)
} else {
arr = append(arr, rand.NormFloat64())
}
}
return;
}
But the memory address is invarient to the increment of size expanding, this really confused me.
By the way, the memory allocation strategy is different in this two implemetation (STL VS. Go), I mean the expanding size. Is there any advantage or disadvantage? Here is the simplified output of code above[size and first element address]:
Golang CPP STL
2 0xc0800386c0 2 004B19C0
4 0xc0800386c0 4 004AE9B8
8 0xc0800386c0 6 004B29E0
16 0xc0800386c0 9 004B2A18
32 0xc0800386c0 13 004B2A68
64 0xc0800386c0 19 004B2AD8
128 0xc0800386c0 28 004B29E0
256 0xc0800386c0 42 004B2AC8
512 0xc0800386c0 63 004B2C20
1024 0xc0800386c0 94 004B2E20
1280 0xc0800386c0 141 004B3118
1600 0xc0800386c0 211 004B29E0
2000 0xc0800386c0 316 004B3080
2500 0xc0800386c0 474 004B3A68
3125 0xc0800386c0 711 004B5FD0
3906 0xc0800386c0 1066 004B7610
4882 0xc0800386c0 1599 004B9768
6102 0xc0800386c0 2398 004BC968
7627 0xc0800386c0 3597 004C1460
9533 0xc0800386c0 5395 004B5FD0
11916 0xc0800386c0 8092 004C0870
14895 0xc0800386c0 12138 004D0558
18618 0xc0800386c0 18207 004E80B0
23272 0xc0800386c0 27310 0050B9B0
29090 0xc0800386c0 40965 004B5FD0
36362 0xc0800386c0 61447 00590048
45452 0xc0800386c0 92170 003B0020
56815 0xc0800386c0 138255 00690020
71018 0xc0800386c0 207382 007A0020
....
UPDATE:
See comments for Golang memory allocation strategy.
For STL, the strategy depends on the implementation. See this post for further information.

Answer 1

Your Go and C++ code fragments are not equivalent. In the C++ function, you are printing the address of the first element in the vector, while in the Go example you are printing the address of the slice itself.
Like a C++ std::vector, a Go slice is a small data type that holds a pointer to an underlying array that holds the data. That data structure has the same address throughout the function. If you want the address of the first element in the slice, you can use the same syntax as in C++: &arr[0].

Answer 2

You're getting the pointer to the slice header, not the actual backing array. You can think of the slice header as a struct like
type SliceHeader struct {
len,cap int
backingArray unsafe.Pointer
}
When you append and the backing array is reallocated, the pointer backingArray will likely be changed (not necessarily, but probably). However, the location of the struct holding the length, cap, and pointer to the backing array doesn't change -- it's still on the stack right where you declared it. Try printing &arr[0] instead of &arr and you should see behavior closer to what you expect.
This is pretty much the same behavior as std::vector, incidentally. Think of a slice as closer to a vector than a magic dynamic array.