Appending to slices in Go - Step-by-Step Execution

Concept Flow - Appending to slices

Start with slice

↓

Call append(slice, value)

↓

Check capacity

↓

Add value

↓

Return new slice with value

↓

End

Start with a slice, append a value, check if capacity allows adding directly or needs new array, then return updated slice.

Execution Sample

Go

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

This code starts with a slice of two integers, appends a third integer, and prints the updated slice.

Execution Table

Step	Action	Slice Before	Value Appended	Capacity Check	Slice After	Output
1	Initialize slice	[]int{1, 2}	-	Capacity 2, length 2	[]int{1, 2}	-
2	Append 3	[]int{1, 2}	3	Capacity full, allocate new array	[]int{1, 2, 3}	-
3	Print slice	[]int{1, 2, 3}	-	-	-	[1 2 3]

💡 Program ends after printing the updated slice with appended value.

Variable Tracker

Variable	Start	After Append	Final
s	[]int{1, 2}	[]int{1, 2, 3}	[]int{1, 2, 3}

Key Moments - 2 Insights

Why does the slice get a new underlying array when appending?

Because the original slice's capacity is full (length equals capacity), Go allocates a new array with more space to hold the appended value, as shown in step 2 of the execution_table.

Does append modify the original slice or return a new one?

Append returns a new slice which may point to a new underlying array. You must assign it back to the variable (s = append(...)) to keep the updated slice, as shown in step 2.

Visual Quiz - 3 Questions

Look at the execution_table at step 2, what happens when appending value 3?

A The value 3 is added directly because there is enough capacity.

B A new underlying array is allocated because capacity is full.

C The slice remains unchanged.

D The program crashes.

💡 Hint

Check the 'Capacity Check' and 'Slice After' columns in step 2 of the execution_table.

According to variable_tracker, what is the value of slice 's' after append?

A []int{1, 2}

B []int{3}

C []int{1, 2, 3}

D []int{}

💡 Hint

Look at the 'After Append' and 'Final' columns for variable 's' in variable_tracker.

If the original slice had capacity 4, what would change in the execution_table at step 2?

A Capacity check would say enough capacity, no new array allocated.

B Capacity check would still allocate new array.

C Slice would be empty after append.

D Append would fail.

💡 Hint

Think about how capacity affects whether a new array is allocated during append, as shown in step 2.

Concept Snapshot

Appending to slices in Go:
- Use append(slice, value) to add elements.
- If capacity allows, value is added directly.
- If capacity is full, Go allocates a new underlying array.
- Always assign the result back to the slice variable.
- Example: s = append(s, 3)

Full Transcript

This visual execution shows how appending to slices in Go works step-by-step. We start with a slice of two integers. When we append a third integer, Go checks if the slice has enough capacity. Since the capacity is full, it allocates a new underlying array with more space and copies the old elements plus the new one. The append function returns this new slice, which we assign back to the variable. Finally, printing the slice shows the updated contents. Key points include understanding capacity, the need to assign the result of append, and how Go manages memory behind the scenes.

Practice

1. What does the append function do in Go when used with slices?

easy

A. It adds new elements to the end of a slice, returning the updated slice.

B. It removes elements from the beginning of a slice.

C. It sorts the elements of a slice in ascending order.

D. It copies elements from one slice to another without changing size.

Solution

1. Step 1: Understand the purpose of append

   The append function is designed to add elements to the end of a slice in Go.

2. Step 2: Recognize the return value

   Append returns a new slice that includes the original elements plus the added ones.

3. Final Answer:

   It adds new elements to the end of a slice, returning the updated slice. -> Option A

4. Quick Check:

   append adds elements to slice end = A [OK]

Hint: Append adds elements at the end and returns the new slice [OK]

Common Mistakes:

• Thinking append removes elements
• Assuming append sorts the slice
• Not realizing append returns a new slice

2. Which of the following is the correct way to append the integer 5 to a slice named nums?

easy

A. nums.append(5)

B. append(nums, 5)

C. nums = append(nums, 5)

D. nums += 5

Solution

1. Step 1: Recall append syntax in Go

   Append returns a new slice, so you must assign it back to the original slice variable.

2. Step 2: Check each option

   nums = append(nums, 5) correctly assigns the result of append back to nums. Options A, B, and D don't work correctly.

3. Final Answer:

   nums = append(nums, 5) -> Option C

4. Quick Check:

   Assign append result back to slice = C [OK]

Hint: Always assign append result back to your slice variable [OK]

Common Mistakes:

• Not assigning append result back to slice
• Using method call syntax like nums.append()
• Trying to add element with += operator

3. What will be the output of this Go code?

nums := []int{1, 2, 3}
nums = append(nums, 4, 5)
fmt.Println(nums)

medium

A. [1 2 3]

B. Compilation error

C. [4 5]

D. [1 2 3 4 5]

Solution

1. Step 1: Understand initial slice and append

   nums starts as [1, 2, 3]. Append adds 4 and 5 to the end.

2. Step 2: Result after append

   After append, nums becomes [1, 2, 3, 4, 5]. Printing nums shows this full slice.

3. Final Answer:

   [1 2 3 4 5] -> Option D

4. Quick Check:

   Appending multiple elements adds all to slice end = D [OK]

Hint: Append can add multiple elements at once [OK]

Common Mistakes:

• Expecting append to replace elements
• Thinking append returns only new elements
• Confusing slice initialization syntax

4. Identify the error in this Go code snippet:

var letters []string
append(letters, "a")
fmt.Println(letters)

medium

A. append result is not assigned back to letters

B. letters slice is nil and cannot be appended to

C. Syntax error in append function call

D. fmt.Println cannot print slices

Solution

1. Step 1: Check slice initialization

   letters is a nil slice but append can still add elements to it.

2. Step 2: Check append usage

   Append returns a new slice; the code does not assign this back to letters, so letters remains nil.

3. Final Answer:

   append result is not assigned back to letters -> Option A

4. Quick Check:

   Always assign append result back to slice = B [OK]

Hint: Assign append result to slice even if slice is nil [OK]

Common Mistakes:

• Assuming append modifies slice in place
• Thinking nil slice cannot be appended to
• Believing fmt.Println can't print slices

5. Given a slice data := []int{1, 2, 3}, which code snippet correctly appends all elements from another slice extra := []int{4, 5, 6} to data?

hard

A. data = append(data, &extra...)

B. data = append(data, extra...)

C. data = append(data, extra)

D. data = append(data, *extra)

Solution

1. Step 1: Understand appending slices

   To append all elements of one slice to another, use the ... operator to unpack elements.

2. Step 2: Analyze options

   data = append(data, extra...) correctly uses append with extra unpacked as extra.... Options B, C, and D have syntax errors or incorrect usage.

3. Final Answer:

   data = append(data, extra...) -> Option B

4. Quick Check:

   Use ... to append all slice elements = A [OK]

Hint: Use ... to unpack slice elements when appending [OK]

Common Mistakes:

• Trying to append slice without ... (adds slice as single element)
• Using pointer or dereference incorrectly
• Trying to append slice as single element