Lab: Arrays

All the code fragments in this lab are complete mini-programs, so you can paste them into the editor and run them to see the results:

Running the code fragments

Right click in Explorer pane to create a new file, e.g. test.go
Paste the question code snippet into the editor pane
Open the terminal window and execute go run test.go
Re-use your test.go file by replacing the content with that of the next question.

Questions

Select the correct statement(s) for initialising an array.
1. var arr [5]string := [5]string {}
2. var arr [5]string = [5]string {}
3. arr := [5]string {}
4. var arr := […]string {}
Reveal

B, C

We can rule out A and D immediately, since you canot use := in any var statement.
What is the length and capacity of [1, 2, 3, 4]
- Length = 4, capacity = 6
- Length = 4, capacity = 10
- Length = Capacity = 4
- Length = 4, capacity = 3
Reveal

Length = Capacity = 4

With no further information about how the list was allocated, then we can assume that both length and capacity are equal to the number of elements in the list.
What would be the output of the following program:
```
package main

import "fmt"

func main() {
        arr := [...]string{1, 2, 3, 4, 5}
        fmt.Println(arr)
}
```
- []
- Error
- [1 2 3 4 5]
- ["1", "2", "3", "4", "5"]
Reveal

Error

The program would not compile. You cannot initialize a string array with integer values.
What would be the output of the following program:
```
package main

import "fmt"

func main() {
        arr := [5]bool{true, true, true, true}
        for i := 0; i < len(arr); i++ {
                if arr[i] {
                        fmt.Println(i)
                }
        }
}
```
- 0
  1
  2
  3
- 0 1 2 3 4
- 0
  1
  2
  3
  4
- Error
Reveal

0
1
2
3
- An array large enough for 5 bool values has been created.
- Values of true have been provided for the first 4 array elements - this means that the remaining one element in the array (arr[4]) will have the default value for bool which is false. So the array looks like this
  
  0 1 2 3 4
  
  true true true true false
- The for loop will iterate across all five values in the array, counting from zero to 4 inclusive.
- Since arr[4] contains false, only numbers as far as 3 will be printed.
What would be the output of the following program:
```
package main

import "fmt"

func main() {
        arr := [5]bool{true, true}
        fmt.Println(arr)
}
```
- [true true false false true]
- [true true]
- [true true false false false]
- [true true true true true]
Reveal

[true true false false false]

Again, this is about knowing that the default value of the array's type will be used to intialize the elements of the array for which explict values have not been provded. This is a 5 element array, and values have only been provided for the first two elements. It looks like this:

0 1 2 3 4

true true false false false
What would be the output of the following program:
```
package main

import "fmt"

func main() {
        arr := [4]int{10, 20, 30, 50, 90}
        fmt.Println(arr)
}
```
- [10 20 30 40 50
- [10 20 30 40 0]
- [10 20 30 40]
- Error
Reveal

Error
- An array large enough for 4 values has been declared.
- The initializer is passing 5 values. This is an error as the values won't fit, so the program will not compile.
What would be the output of the following program:
```
package main

import "fmt"

func main() {
        arr := [10]int{10, 20, 30, 50}
        fmt.Println(arr)
        fmt.Println(len(arr))
}
```
- [10 20 30 50]
- [10 20 30 50 0 0 0 0 0 0]
  4
- [10 20 30 50 0 0 0 0 0 0]
  10
- [10 20 30 50]
  10
Reveal

[10 20 30 50 0 0 0 0 0 0]
10
- The array has been declared to have 10 elements - this means len(arr) is 10.
- It has been initialized with 4 values, therefore the remaining 6 values will have the default value for int which is zero. It looks like this
  
  0 1 2 3 4 5 6 7 8 9
  
  10 20 30 50 0 0 0 0 0 0
What would be the output of the following program:
```
package main

import "fmt"

func main() {
        arr := [10]int{10, 20, 30, 50}
        fmt.Println(arr[0])
        fmt.Println(arr[2])
        fmt.Println(arr[4])
        fmt.Println(arr[8])
        fmt.Println(arr[10])
}
```
- Error
- 10 30 0 0 0
  10
- 10
  30
  0
  0
  nil
- 10
  30
  0
  0
  0
Reveal

Error
- Remember that array indexes start at zero, not 1.
- The array has been declared to have 10 elements. These will be arr[0] thru arr[9]
- arr[10] is trying to access the 11th element which doesn't exist, therefore the program will not compile.
What would be the output of the following program:
```
package main

import "fmt"

func main() {
        arr := [5]string{"a", "b", "c"}
        for index, element := range arr {
                fmt.Println(index, "->", element)
        }
}
```
- 0 -> a
  1 -> b
  2 -> c
  3 -> d
  4 -> e
- 0 -> a
  1 -> b
  2 -> c
  3 ->
  4 ->
- 0 -> a
  1 -> b
  2 -> c
  3 -> 3
  4 -> 4
- Error
Reveal

0 -> a
1 -> b
2 -> c
3 ->
4 ->
- We have a 5 element string array, for which only the first 3 values have been initialized. The remaining 2 values will have the default value for string, which is an empty string ("") It looks like this
  
  0 1 2 3 4
  
  a b c
- The range keyword returns the index and value for each element in the array.
- Thus 3 and 4 print as they do, because their values are empty.
What would be the output of the following program:
```
package main

import "fmt"

func main() {
        arr := [5][2]string{{"a"}, {"b"}, {"c"}}
        fmt.Println(arr[0][0])
        fmt.Println(arr[1][1])
        fmt.Println(arr[2][0])
}
```
- a
  
  c
- a
  b
  c
- a
  c
- Error
Reveal

a

c

Here we have a two dimensional array, 5 wide and 2 deep. And there are default values everywhere! Let's visualize this data structure. It's like an Excel spreadsheet!

0 1 2 3 4

0 a b c

1

Given arr[x][y], look up x as the column number and y as the row number.

Thus arr[0][0] is a, arr[1][1] is empty, arr[2][0] is c.

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