Common pointer use cases in Go - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a pointer to an integer variable.

Go

var x int = 10
var p [1] = &x

Drag options to blanks, or click blank then click option'

A*int

Bint*

C&int

Dint

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'int' instead of '*int' for pointer type.

Using '&int' which is invalid syntax.

Using 'int*' which is not Go syntax.

In Go, a pointer to an int is declared as *int. The & operator gets the address of x.

2fill in blank

medium

Complete the code to change the value of an integer through its pointer.

Go

var x int = 5
var p *int = &x
*p [1] 20

Drag options to blanks, or click blank then click option'

A=

B==

C:=

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '==' which is a comparison, not assignment.

Using ':=' which declares a new variable, not assignment.

Using '!=' which is a comparison operator.

To assign a new value through a pointer, use =. The expression *p = 20 changes the value of x to 20.

3fill in blank

hard

Fix the error in the code to correctly swap two integers using pointers.

Go

func swap(a, b [1] int) {
    temp := *a
    *a = *b
    *b = temp
}

Drag options to blanks, or click blank then click option'

A**

B&&

C&

D*

Attempts:

3 left

💡 Hint

Common Mistakes

Using '&int' which is invalid syntax.

Using '**int' which is a pointer to a pointer, not needed here.

Using '&&int' which is invalid.

The function parameters should be pointers to int, declared as *int. This allows swapping the values at the addresses.

4fill in blank

hard

Fill both blanks to create a map where keys are strings and values are pointers to integers.

Go

m := map[string][1]{}
m["score"] = [2]

Drag options to blanks, or click blank then click option'

A*int

B&x

Cint

Dx

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'int' as map value type instead of '*int'.

Assigning 'x' instead of '&x' to the map value.

Using '&int' which is invalid.

The map values should be pointers to int, so *int. To assign a pointer, use the address operator & on a variable like x.

5fill in blank

hard

Fill all three blanks to define a struct with a pointer field, create an instance, and assign a value through the pointer.

Go

type Person struct {
    age [1]
}
p := Person{age: nil}
var a int = 30
p.age = [2]
*[3] = 35

Drag options to blanks, or click blank then click option'

A*int

B&a

Cp.age

Dint

Attempts:

3 left

💡 Hint

Common Mistakes

Declaring age as int instead of *int.

Assigning 'a' instead of '&a' to p.age.

Dereferencing wrong variable instead of p.age.

The struct field age is a pointer to int (*int). Assign the address of a to p.age. Then dereference p.age to change the value.

Practice

1. What does a pointer in Go store?

easy

A. The memory address of a variable

B. The value of a variable

C. The type of a variable

D. The size of a variable

Solution

1. Step 1: Understand what a pointer represents

   A pointer holds the location in memory where a variable is stored, not the value itself.

2. Step 2: Differentiate pointer from value

   The value is stored at the address the pointer points to, not inside the pointer variable.

3. Final Answer:

   The memory address of a variable -> Option A

4. Quick Check:

   Pointers store addresses = B [OK]

Hint: Pointers hold addresses, not values directly [OK]

Common Mistakes:

• Confusing pointer with the actual value
• Thinking pointer stores variable type
• Assuming pointer stores variable size

2. Which of the following is the correct way to declare a pointer to an int in Go?

easy

A. var p int*

B. var p &int

C. var p *int

D. var *p int

Solution

1. Step 1: Recall Go pointer syntax

   In Go, a pointer type is written as *Type, so a pointer to int is *int.

2. Step 2: Check variable declaration format

   The variable name comes after var, then the type. So correct is var p *int.

3. Final Answer:

   var p *int -> Option C

4. Quick Check:

   Pointer syntax is var name *Type = D [OK]

Hint: Pointer type uses * before type, not after variable [OK]

Common Mistakes:

• Placing * after variable name
• Using & instead of * for pointer type
• Incorrect order of * and variable name

3. What is the output of this Go code?

func main() {
    x := 10
    p := &x
    *p = 20
    fmt.Println(x)
}

medium

A. 10

B. Memory address of x

C. Compilation error

D. 20

Solution

1. Step 1: Understand pointer assignment

   p points to x's address. Changing *p changes the value at x's address.

2. Step 2: Trace value change

   *p = 20 sets x to 20. So printing x outputs 20.

3. Final Answer:

   20 -> Option D

4. Quick Check:

   Pointer changes original value = 20 [OK]

Hint: Changing *p changes original variable value [OK]

Common Mistakes:

• Thinking x remains 10 after *p = 20
• Printing pointer address instead of value
• Assuming compilation error due to pointer

4. Identify the error in this Go code snippet:

func updateValue(p *int) {
    *p = 100
}

func main() {
    var p *int
    updateValue(p)
}

medium

A. Syntax error in function declaration

B. p is nil and dereferenced causing runtime panic

C. updateValue should return an int

D. No error, code runs fine

Solution

1. Step 1: Check pointer initialization

   p is declared but not assigned an address, so it is nil.

2. Step 2: Analyze dereferencing nil pointer

   updateValue dereferences *p, causing a runtime panic because p is nil.

3. Final Answer:

   p is nil and dereferenced causing runtime panic -> Option B

4. Quick Check:

   Dereferencing nil pointer causes panic = A [OK]

Hint: Never dereference nil pointers without initialization [OK]

Common Mistakes:

• Assuming nil pointer is safe to dereference
• Thinking function must return a value
• Ignoring runtime panic possibility

5. Given this Go code, what will be the output?

func swap(a, b *int) {
    *a, *b = *b, *a
}

func main() {
    x, y := 5, 10
    swap(&x, &y)
    fmt.Println(x, y)
}

hard

A. 10 5

B. 5 10

C. Memory addresses of x and y

D. Compilation error due to swap syntax

Solution

1. Step 1: Understand swap function with pointers

   swap receives pointers to x and y, then swaps the values they point to.

2. Step 2: Trace swap effect on variables

   After swap, x becomes 10 and y becomes 5.

3. Final Answer:

   10 5 -> Option A

4. Quick Check:

   Pointer swap changes original values = 10 5 [OK]

Hint: Swap values by swapping *a and *b via pointers [OK]

Common Mistakes:

• Expecting original values unchanged
• Printing pointer addresses instead of values
• Thinking swap function syntax is invalid