Slicing Quiz

What does this code print?

nums := []int{9, 7, 5}
nums = append(nums, 2, 4, 6)

fmt.Println(nums[2:4])

[9 7 5 2 4 6]
[5 2] CORRECT
[4 6]
[7 2]
[9 7]

2: nums is [9 7 5 2 4 6]. So, nums[2:4] is [5 2]. Remember, in nums[2:4] -> 2 is the starting index, so nums[2] is 5; And 4 is the stopping position, so nums[4-1] is 2 (-1 because the stopping position is the element position). So, nums[2:4] returns a new slice that contains the elements at the middle of the nums slice.

What does this code print?

nums := []int{9, 7, 5}
nums = append(nums, 2, 4, 6)

fmt.Println(nums[:2])

[9 7 5 2 4 6]
[5 2]
[4 6]
[7 2]
[9 7] CORRECT

5: nums is [9 7 5 2 4 6]. So, nums[:2] is nums[0:2] which in turn returns [9 7].

What does this code print?

nums := []int{9, 7, 5}
nums = append(nums, 2, 4, 6)

fmt.Println(nums[len(nums)-2:])

[9 7 5 2 4 6]
[5 2]
[4 6] CORRECT
[7 2]
[9 7]

3: nums is [9 7 5 2 4 6]. So, nums[len(nums)-2:] is nums[4:6] (len(nums) is 6) which in turn returns [4 6].

What does this code print?

names := []string{"einstein", "rosen", "newton"}
names = names[:]
fmt.Println(names[:1])

[einstein rosen newton]
[einstein rosen]
[einstein] CORRECT
[]

3: names[:] is names[0:3] -> [einstein rosen newton]. names[:1] is names[0:1] -> [einstein].

What is the type of the marked expression below?

names := []string{"einstein", "rosen", "newton"}
names[2:3] // <- marked

[]string CORRECT
string
names
[]int

1: Yes! A slicing expression returns a slice.

2: Remember, a slicing expression returns a slice. Did I give you the answer? Oops.

What is the type of the marked expression below?

names := []string{"einstein", "rosen", "newton"}
names[2] // <- marked

[]string
string CORRECT
names
[]int

1: Remember, an index expression returns an element value, not a slice.

2: Yep! An index expression returns an element value. The element type of the []string slice is string, so the returned value is a string value.

Which index expression returns the "rosen" element?

names := []string{"einstein", "rosen", "newton"}
names = names[1:len(names) - 1]

names[0] CORRECT
names[1]
names[2]

1: That's right: names2 is ["rosen"] after the slicing.

2: That's not right. Remember, indexes are relative to a slice. names is ["einstein" "rosen" "newton"] but names[1:len(names)-1] is ["rosen"]. So, names2[1] is an error, it's because, the length of the last slice is 1.

What does this code print?

names := []string{"einstein", "rosen", "newton"}
names = names[1:]
names = names[1:]
fmt.Println(names)

[einstein rosen newton]
[rosen newton]
[newton] CORRECT
[]

3: Remember, slicing returns a new slice. Here, each names = names[1:] statement overwrites the names slice with the newly returned slice from the slicing. At first, the names was [einstein rosen newton]. After the first slicing, the names becomes [rosen newton]. After the second slicing, names becomes [newton]. See this for the complete explanation: https://play.golang.org/p/EsEHrSeByFR

What does this code print?

i := 2
s := fmt.Sprintf("i = %d * %d = %d", i, i, i*i)
fmt.Print(s)

i = i * i = i*i
i = %d * %d = %d
i = 2 * 2 = 2
i = 2 * 2 = 4 CORRECT

4: Awesome! Sprintf works just like Printf. Instead of printing the result to standard out (usually to command-line prompt), it returns a string value.