Slice creation in Go - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of slice length and capacity

What is the output of this Go program?

Go

package main
import "fmt"
func main() {
    s := make([]int, 3, 5)
    fmt.Println(len(s), cap(s))
}

A 5 3

B 3 5

C 0 5

D 3 3

💡 Hint

Remember: len is the number of elements, cap is the total allocated space.

The slice is created with length 3 and capacity 5, so len(s) is 3 and cap(s) is 5.

❓ Predict Output

intermediate

Slice literal with capacity

What will this program print?

Go

package main
import "fmt"
func main() {
    s := []int{1, 2, 3, 4}
    fmt.Println(len(s), cap(s))
}

A 3 4

B 4 8

C 4 4

D 4 0

💡 Hint

Slice literals have capacity equal to their length.

A slice literal creates a slice with length and capacity equal to the number of elements.

❓ Predict Output

advanced

Appending beyond capacity

What is the output of this Go program?

Go

package main
import "fmt"
func main() {
    s := make([]int, 2, 3)
    s[0], s[1] = 10, 20
    s = append(s, 30)
    s = append(s, 40)
    fmt.Println(len(s), cap(s))
}

A 4 6

B 4 3

C 4 4

D 3 3

💡 Hint

Appending beyond capacity causes the slice to grow, usually doubling capacity.

The initial capacity is 3. After appending the 4th element, capacity grows, typically doubling to 6.

❓ Predict Output

advanced

Slice zero value behavior

What happens when you run this program?

Go

package main
import "fmt"
func main() {
    var s []int
    fmt.Println(len(s), cap(s))
    s = append(s, 1)
    fmt.Println(len(s), cap(s))
}

A

0 1
1 1

B

0 0
1 0

C

1 1
1 1

D

0 0
1 1

💡 Hint

A nil slice has zero length and capacity but can be appended to.

Initially, s is nil with length and capacity 0. After append, length and capacity become 1.

❓ Predict Output

expert

Slice header and underlying array

What will this program print?

Go

package main
import "fmt"
func main() {
    arr := [4]int{1, 2, 3, 4}
    s := arr[1:3]
    s = append(s, 5)
    fmt.Println(arr)
    fmt.Println(s)
}

A

[1 2 3 5]
[2 3 5]

B

[1 2 3 4]
[2 3 5]

C

[1 2 3 4]
[2 3 4 5]

D

[1 2 3 4]
[2 3 4]

💡 Hint

Appending within capacity modifies the shared underlying array.

s := arr[1:3] has length 2, capacity 3 (arr[1:4]). Appending 5 increases length to 3 within capacity, so it overwrites arr[3] with 5. Now arr is [1 2 3 5] and s is [2 3 5].