Is it possible to slice an inner matrix with brackets? - go

Question

I am modifying the perimeter values on a matrix, then trying to recurse into the inner values. I expected I'd be able to access the inner values with something like matrix[1:3][1:3]. This is not the case, and I'm a bit lost as to the underlying logic of how Go handles sequential brackets.
package main
import (
"fmt"
)
var m = [][]int{
[]int{0, 1, 2, 3},
[]int{4, 5, 6, 7},
[]int{8, 9, 10, 11},
[]int{12, 13, 14, 15},
}
I am trying to access the values 5, 6, 9, 10 in the above matrix -- the "inner" matrix.
func main() {
inner := m[1:3][1:3]
fmt.Printf("%#v\n", inner)
// Expected Output: [][]int{
// []int{5, 6},
// []int{9, 10}
// }
// Actual Ouput: [][]int{
// []int{8, 9, 10, 11},
// []int{12, 13, 14, 15}
// }
inner = m[1:3]
fmt.Printf("%#v\n", inner)
// Output:
// [][]int{
// []int{4, 5, 6, 7},
// []int{8, 9, 10, 11}
// }
inner = innerMatrix(m)
fmt.Printf("%#v\n", inner)
// [][]int{
// []int{5, 6},
// []int{9, 10}
}
func innerMatrix(m [][]int) (inner [][]int) {
innerRows := m[1:3]
for _, row := range innerRows {
inner = append(inner, row[1:3])
}
return
}
The function innerMatrix is able to produce the output I expected. I'm lost as to why (1) m[1:3][1:3] doesn't have the same effect, (2) it seems to translate to m[2:4]. What's going on?
Playground

Answer 1

When you create a sub-slice in Go, you're allowed to grow that sub-slice back to the capacity of the original slice again, e.g.:
package main
import "fmt"
func main() {
a := []int{1, 2, 3, 4, 5}
b := a[1:3]
fmt.Printf("a[1:3]: %v\n", b)
c := b[1:3]
fmt.Printf("b[1:3]: %v\n", c)
}
which outputs:
a[1:3]: [2 3]
b[1:3]: [3 4]
Note that b only has two elements, but we're allowed to create a slice of the second and third elements because the capacity of the slice it's a sub-slice of is sufficiently large, and all the slices share the same underlying array. See the last example in section "Slice internals" on this page
So what's happening in your case is that m[1:3] is equivalent to:
var m1 = [][]int{
[]int{4, 5, 6, 7}, // second element of m
[]int{8, 9, 10, 11}, // third element of m
}
and m[1:3][1:3] is therefore equivalent to m1[1:3], which is equivalent to:
var m2 = [][]int{
[]int{8, 9, 10, 11}, // second element of m1
[]int{12, 13, 14, 15}, // "third" element of m1
}
with the "third" element appearing only because the capacity of m is large enough to contain it, and does in fact contain it. If m only had three elements, this would cause a panic.
In other words, m[1:3][1:3] is here exactly equivalent to m[2:4], because m[1:3][1:3] gives you the second and third elements of m[1:3]. Maybe easier to understand with a diagram:
m : []int{1, 2, 3, 4}
m[1:3] : []int{ 2, 3 }
m[1:3][1:3] : []int{ 3, 4}
m[2:4] : []int{ 3, 4}
As an oversimplification, you can imagine that the square brackets give the requested elements of whatever is immediately to the left of them, so a somewhat extreme example:
package main
import "fmt"
func main() {
a := []int{1, 2, 3, 4, 5}
b := a[1:5]
fmt.Printf("b: %v, a[1:5] : %v\n",
b, a[1:5])
c := b[1:4]
fmt.Printf("c: %v , a[1:5][1:4] : %v\n",
c, a[1:5][1:4])
d := c[1:3]
fmt.Printf("d: %v , a[1:5][1:4][1:3] : %v\n",
d, a[1:5][1:4][1:3])
e := d[1:2]
fmt.Printf("e: %v , a[1:5][1:4][1:3][1:2]: %v\n",
e, a[1:5][1:4][1:3][1:2])
}
which outputs:
b: [2 3 4 5], a[1:5] : [2 3 4 5]
c: [3 4 5] , a[1:5][1:4] : [3 4 5]
d: [4 5] , a[1:5][1:4][1:3] : [4 5]
e: [5] , a[1:5][1:4][1:3][1:2]: [5]

Answer 2

Here is your data:
var m = [][]int{
[]int{0, 1, 2, 3},
[]int{4, 5, 6, 7},
[]int{8, 9, 10, 11},
[]int{12, 13, 14, 15},
}
First you ask what is inner := m[1:3][1:3]?
Well, taking one at a time, m[1:3] is the subslice you get when grabbing elements 1 to 3 not including 3. So that is elements 1 and 2. Element 0 is []int{0, 1, 2, 3}, element 1 is []int{4, 5, 6, 7}, and element 2 is []int{8, 9, 10, 11},
So m[1:3] is
[][]int{
[]int{4, 5, 6, 7},
[]int{8, 9, 10, 11},
}
Now, to get m[1:3][1:3] we repeat the same thing on this result.
However, this time, there are only two elements. We again want elements 1 and 2. We skip element 0 which is []int{4, 5, 6, 7}. Element 1 is []int{8, 9, 10, 11}. There is no element 2. However, the result of m[1:3] was a subslice and the underlying slice had an additional element.
So by extending this subslice we can get that element back again, and that is the hiddent element 2 which is []int{12, 13, 14, 15}.
So m[1:3][1:3] is
[][]int{
[]int{8, 9, 10, 11},
[[]int{12, 13, 14, 15},
}
Finally, you wonder why func innerMatrix works differently? It starts by producing m[1:3] which is
[][]int{
[]int{4, 5, 6, 7},
[]int{8, 9, 10, 11},
}
Then the code IS NOT taking another subslice in the same way that m[1:3][1:3] is two consecutive subslices. Instead, you take each element in the slice and grab the subslice from each element. The first time through you grab subslices [1:3] from []int{4, 5, 6, 7}. The result of that is
[]int{5, 6}
The second time through you do the same on the second element []int{8, 9, 10, 11} which is
[]int{9, 10}
Finally, each time through you append the result to the result slice, so you append []int{5, 6} to [], to give [][]{int{5, 6}}, and then you append []int{9, 10} to give
[][]{int{5, 6}, []int{9, 10} }