Method call behavior in Go - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of method call with pointer vs value receiver

What is the output of this Go program?

Go

package main
import "fmt"
type Counter struct {
    count int
}
func (c Counter) Increment() {
    c.count++
}
func (c *Counter) IncrementPtr() {
    c.count++
}
func main() {
    c := Counter{count: 5}
    c.Increment()
    fmt.Println(c.count)
    c.IncrementPtr()
    fmt.Println(c.count)
}

A

6
5

B

5
5

C

5
6

D

6
6

💡 Hint

Think about whether the method receiver is a value or a pointer and how it affects the original struct.

The method Increment has a value receiver, so it works on a copy of the struct and does not change the original count. The method IncrementPtr has a pointer receiver, so it modifies the original struct's count. Therefore, the first print shows 5, and the second shows 6.

❓ Predict Output

intermediate

Method call on nil pointer receiver

What will this Go program print?

Go

package main
import "fmt"
type Node struct {
    value int
    next  *Node
}
func (n *Node) PrintValue() {
    if n == nil {
        fmt.Println("nil node")
        return
    }
    fmt.Println(n.value)
}
func main() {
    var n *Node = nil
    n.PrintValue()
}

A compilation error

B 0

C runtime error: invalid memory address or nil pointer dereference

D nil node

💡 Hint

Check how method calls on nil pointers behave when the method has a pointer receiver.

In Go, calling a method with a pointer receiver on a nil pointer is allowed as long as the method itself does not dereference the pointer without checking. Here, the method checks if n == nil before accessing n.value, so it prints "nil node" safely.

🔧 Debug

advanced

Why does this method call cause a compilation error?

Consider this Go code snippet. Why does calling p.Increment() cause a compilation error?

Go

package main
import "fmt"
type Point struct {
    x, y int
}
func (p *Point) Increment() {
    p.x++
    p.y++
}
func main() {
    var p Point
    p.Increment()
    fmt.Println(p)
}

A No error; the code compiles and prints {1 1}

B Compilation error because method Increment has pointer receiver but called on value without address

C Runtime panic due to nil pointer dereference

D Compilation error because method Increment is not defined for type Point

💡 Hint

Think about how Go handles method calls with pointer receivers on values.

In Go, methods with pointer receivers can be called on values because the compiler automatically takes the address. So this code actually compiles and runs fine, printing {1 1}. Therefore, option A is correct.

❓ Predict Output

advanced

Output of interface method call with pointer and value receivers

What is the output of this Go program?

Go

package main
import "fmt"
type Printer interface {
    Print()
}
type Data struct {
    value int
}
func (d Data) Print() {
    fmt.Println("Value:", d.value)
}
func (d *Data) PrintPtr() {
    fmt.Println("Pointer Value:", d.value)
}
func main() {
    d := Data{value: 10}
    var p Printer = d
    p.Print()
    var p2 Printer = &d
    p2.Print()
}

A

Value: 10
Pointer Value: 10

B

Value: 10
Value: 10

C

Pointer Value: 10
Pointer Value: 10

D Compilation error: cannot assign &d to Printer

💡 Hint

Check which methods satisfy the interface for value and pointer types.

The interface Printer requires a Print() method. The value type Data has a Print() method, so both d and &d implement Printer. The method PrintPtr() is not part of the interface, so it is not called. Therefore, both calls print "Value: 10".

🧠 Conceptual

expert

Method sets and interface implementation rules

Which statement about method sets and interface implementation in Go is correct?

A A pointer to type T implements an interface if all interface methods have pointer receivers or value receivers on T

B A pointer to type T implements an interface if all interface methods have value receivers on T

C A value of type T implements an interface if all interface methods have pointer receivers on T

D A value of type T implements an interface only if all interface methods have pointer receivers on T

💡 Hint

Recall how Go defines method sets for values and pointers.

In Go, the method set of a value type T includes all methods with value receivers. The method set of *T (pointer to T) includes all methods with pointer or value receivers. Therefore, a pointer to T implements an interface if the interface methods are implemented with either pointer or value receivers. Option A is correct.

Option A is wrong because values do not have pointer receiver methods in their method set. Option A is wrong because pointer types have all methods, not just value receivers. Option A is wrong for the same reason as A.