Address and dereference operators in Go - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of pointer dereference after assignment

What is the output of this Go program?

Go

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

A Compilation error

B 10

C 0

D 20

💡 Hint

Remember that *p changes the value at the address p points to.

The pointer p points to x. Changing *p changes x's value to 20. So printing x outputs 20.

❓ Predict Output

intermediate

Value of variable after pointer reassignment

What will be printed by this Go code?

Go

package main
import "fmt"
func main() {
    a := 5
    b := 10
    p := &a
    p = &b
    *p = 15
    fmt.Println(a, b)
}

A 5 15

B 5 10

C 15 10

D Compilation error

💡 Hint

Pointer p is first set to a, then to b. Changing *p affects which variable?

Initially p points to a, then it points to b. Changing *p = 15 changes b to 15. a remains 5.

🔧 Debug

advanced

Identify the error in pointer usage

What error does this Go code produce?

Go

package main
func main() {
    var p *int
    *p = 10
}

A No error, program prints 10

B Compilation error: cannot assign to *p

C runtime error: invalid memory address or nil pointer dereference

D Compilation error: missing variable declaration

💡 Hint

p is declared but not initialized to point to valid memory.

p is a nil pointer. Dereferencing it (*p) causes a runtime panic due to invalid memory access.

🧠 Conceptual

advanced

Effect of pointer on function argument

What will be the output of this Go program?

Go

package main
import "fmt"
func increment(n *int) {
    *n = *n + 1
}
func main() {
    x := 7
    increment(&x)
    fmt.Println(x)
}

A 7

B 8

C 0

D Compilation error: cannot use &x

💡 Hint

The function changes the value at the address passed.

increment receives a pointer to x and increases the value at that address by 1, so x becomes 8.

❓ Predict Output

expert

Pointer arithmetic and dereference behavior

What is the output of this Go program?

Go

package main
import "fmt"
func main() {
    arr := [3]int{1, 2, 3}
    p := &arr[0]
    fmt.Println(*p)
    p = &arr[1]
    fmt.Println(*p)
    p++
    fmt.Println(*p)
}

A Compilation error: invalid operation p++

B

1
2
Compilation error

C

1
2
0

D

1
2
3

💡 Hint

Go does not support pointer arithmetic like p++.

In Go, you cannot increment pointers with ++. This causes a compilation error.

Practice

1. What does the & operator do in Go?

easy

A. It declares a new variable.

B. It returns the memory address of a variable.

C. It multiplies two numbers.

D. It dereferences a pointer to get the value.

Solution

1. Step 1: Understand the & operator

   The & operator is used to get the address of a variable in memory.

2. Step 2: Differentiate from * operator

   The * operator is used to access the value at an address, not to get the address itself.

3. Final Answer:

   It returns the memory address of a variable. -> Option B

4. Quick Check:

   & operator = address retrieval [OK]

Hint: Remember & gets address, * gets value at address [OK]

Common Mistakes:

• Confusing & with * operator
• Thinking & multiplies values
• Assuming & declares variables

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

easy

A. var p &int = num

B. var p int = *num

C. var p *int = &num

D. var p *num = &int

Solution

1. Step 1: Understand pointer declaration syntax

   In Go, a pointer to an int is declared as *int. To assign it, use the address operator & on the variable.

2. Step 2: Check each option

   var p *int = &num correctly declares p as a pointer to int and assigns the address of num. Others misuse types or operators.

3. Final Answer:

   var p *int = &num -> Option C

4. Quick Check:

   Pointer declaration = var p *int = &num [OK]

Hint: Pointer syntax: var p *Type = &variable [OK]

Common Mistakes:

• Using * on variable instead of type
• Using & with type instead of variable
• Mixing variable and type names

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

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

medium

A. 10

B. Compilation error

C. Memory address of x

D. 20

Solution

1. Step 1: Understand pointer assignment

   Variable p stores the address of x. Using *p = 20 changes the value at that address.

2. Step 2: Check final value of x

   Since *p points to x, changing *p to 20 updates x to 20.

3. Final Answer:

   20 -> Option D

4. Quick Check:

   Dereferencing pointer changes original variable [OK]

Hint: Changing *p changes x because p points to x [OK]

Common Mistakes:

• Thinking *p = 20 creates new variable
• Expecting print to show address
• Confusing & and * operators

4. Identify the error in this Go code snippet:

func main() {
  var a int = 5
  var p *int
  *p = a
  fmt.Println(*p)
}

medium

A. Pointer p is not initialized before dereferencing.

B. Variable a should be a pointer.

C. Cannot assign int to *int directly.

D. fmt.Println cannot print pointer values.

Solution

1. Step 1: Check pointer initialization

   Pointer p is declared but not assigned an address before dereferencing with *p = a.

2. Step 2: Understand consequences

   Dereferencing an uninitialized pointer causes a runtime panic (nil pointer dereference).

3. Final Answer:

   Pointer p is not initialized before dereferencing. -> Option A

4. Quick Check:

   Dereference only initialized pointers [OK]

Hint: Always assign pointer before dereferencing it [OK]

Common Mistakes:

• Dereferencing nil pointer
• Confusing pointer and value assignment
• Assuming fmt.Println can't print pointers

5. Given the code below, what will be the output?

package main
import "fmt"
func updateValue(p *int) {
  *p = *p + 5
}
func main() {
  val := 10
  updateValue(&val)
  fmt.Println(val)
}

hard

A. 15

B. 10

C. Memory address of val

D. Compilation error

Solution

1. Step 1: Understand function parameter

   The function updateValue takes a pointer to an int. Passing &val sends the address of val.

2. Step 2: Analyze pointer dereference and update

   Inside the function, *p = *p + 5 adds 5 to the value at the address, updating val to 15.

3. Final Answer:

   15 -> Option A

4. Quick Check:

   Pointer parameter updates original variable [OK]

Hint: Passing &var lets function modify original value [OK]

Common Mistakes:

• Expecting original val unchanged
• Confusing pointer with value copy
• Thinking function returns new value