How much memory do golang maps reserve?

Question

Given a map allocation where the initial space is not specified, for example:

foo := make(map[string]int)

The documentation suggests that the memory allocation here is implementation dependent. So (how) can I tell how much memory my implementation is allocating to this map?

Answer 1

You may use the Go testing tool to measure size of arbitrary complex data structures. This is detailed in this answer: How to get variable memory size of variable in golang?

To measure the size of a map created by make(map[string]int), use the following benchmark function:

var x map[string]int

func BenchmarkEmptyMap(b *testing.B) {
    for i := 0; i < b.N; i++ {
        x = make(map[string]int)
    }
}

Executing with

go test -bench . -benchmem

The result is:

BenchmarkEmptyMap-4     20000000   110 ns/op      48 B/op    1 allocs/op

So the answer is on my 64-bit architecture: 48 bytes.

As hinted, size may depend on architecture. Also size may depend on the initial capacity you may pass to make(), as you can see in this example:

func BenchmarkEmptyMapCap100(b *testing.B) {
    for i := 0; i < b.N; i++ {
        x = make(map[string]int, 100)
    }
}

Output:

BenchmarkEmptyMapCap100-4   1000000    1783 ns/op   4176 B/op    3 allocs/op

A map of type map[string]int with an initial capacity of 100 now requires 4176 bytes (on 64-bit arch).

The default initial capacity is around 7 if not specified explicitly.

Answer 2

If you look at the source of Go's map type, you will see, that a map consists of a header (type hmap) and an array of buckets (type bmap). When you create a new map and don't specify the initial space (hint), only one bucket is created.

A header consists of several fields:

1 * int,
2 * uint8,
1 * uint16,
1 * uint32,
2 * unsafe.Pointer,
1 * uintptr.

Size of the types int, uintptr, and unsafe.Pointer equals the size of a word (8 bytes on 64 bit machines).

A bucket consists of an array of 8 * uint8.

This gives a total of 40 + 8 = 48 bytes (64 bit architecture)