Slice of slices types

Question

I'm currently working my way through the excellent Tour of Go. I finished one of the exercises (#45) with the following solution:

func Pic(dx, dy int) [][]uint8 {     pic := make([][]uint8, dy) /* type declaration */     for i := range pic {         pic[i] = make([]uint8, dx) /* again the type? */         for j := range pic[i] {             pic[i][j] = uint8((i+j)/2)         }     }     return pic } 

I don't understand why I have to use a make statement with the uint8 type twice (see comments in snippet). That seems redundant but I can't figure out how to do it in an other way.

Answer 1

To be explicit, we can use parentheses to rewrite [][]uint8 as []([]uint8): a slice of (slices of type uint8).

Using the make built-in function, for a slice of type T, make(T, n) returns a slice of type T with length n and capacity n.

Therefore, make([][]uint8, 2) is equivalent to make([]([]uint8), 2), it returns a slice, with length and capacity of 2, of slices of type uint8, where each slice of type uint8 is initialized to its zero value (a nil reference with a length and capacity of zero).

Multi-dimensional slices are jagged and are analogous to multi-dimensional jagged arrays.

For example,

package main  import "fmt"  func main() {     ss := make([][]uint8, 2) // ss is []([]uint8)     fmt.Printf("ss:    %T %v %d\n", ss, ss, len(ss))     for i, s := range ss { // s is []uint8         fmt.Printf("ss[%d]: %T %v %d\n", i, s, s, len(s))     } } 

Output:

ss:    [][]uint8 [[] []] 2 ss[0]: []uint8 [] 0 ss[1]: []uint8 [] 0