Passing values vs pointers in Go

Introduction

Passing values copies the data, while passing pointers shares the original data. This helps control if changes affect the original or just a copy.

When you want to keep the original data safe from changes.

When you want a function to update the original data.

When you want to save memory by not copying large data.

When you want to share data between parts of your program.

When you want to avoid slow copying of big structures.

Syntax

Go

func functionName(valueType value) {
    // code using value
}

func functionName(pointerType *valueType) {
    // code using pointer
}

Passing by value copies the data into the function.

Passing by pointer sends the address, so the function can change the original data.

Examples

First function changes only its copy. Second function changes the original variable through the pointer.

Go

func updateValue(x int) {
    x = 10
}

func updatePointer(x *int) {
    *x = 10
}

Shows how passing value does not change original, but passing pointer does.

Go

var a int = 5
updateValue(a)
// a is still 5

updatePointer(&a)
// a is now 10

Sample Program

This program shows the difference between passing a value and passing a pointer. The first function call does not change the original variable. The second function call changes it.

Go

package main

import "fmt"

func changeValue(val int) {
    val = 100
}

func changePointer(ptr *int) {
    *ptr = 100
}

func main() {
    a := 50
    fmt.Println("Before changeValue:", a)
    changeValue(a)
    fmt.Println("After changeValue:", a)

    fmt.Println("Before changePointer:", a)
    changePointer(&a)
    fmt.Println("After changePointer:", a)
}

Important Notes

Use pointers when you want to modify the original data inside a function.

Passing large structs by pointer can improve performance by avoiding copying.

Be careful with pointers to avoid unexpected changes or nil pointer errors.

Summary

Passing by value copies data; changes inside function do not affect original.

Passing by pointer shares the original data; changes inside function affect original.

Choose based on whether you want to protect or modify the original data.

Practice

1. What happens when you pass a variable to a function by value in Go?

easy

A. The function receives a copy; changes inside do not affect the original variable.

B. The function receives the original variable; changes inside affect the original.

C. The function cannot modify the variable at all.

D. The function receives a reference but cannot change the original variable.

Solution

1. Step 1: Understand passing by value

   Passing by value means the function gets a copy of the variable, not the original.

2. Step 2: Effect of changes inside the function

   Changes to the copy do not affect the original variable outside the function.

3. Final Answer:

   The function receives a copy; changes inside do not affect the original variable. -> Option A

4. Quick Check:

   Passing by value = copy, no original change [OK]

Hint: Passing by value means function works on a copy [OK]

Common Mistakes:

• Thinking changes inside function affect original variable
• Confusing pointers with values
• Assuming function cannot modify any data

2. Which of the following is the correct way to declare a function parameter as a pointer in Go?

easy

A. func updateValue(val *int) {}

B. func updateValue(&val int) {}

C. func updateValue(val int) {}

D. func updateValue(*val int) {}

Solution

1. Step 1: Recall pointer syntax in Go function parameters

   In Go, a pointer parameter is declared by placing * before the type, e.g., *int.

2. Step 2: Identify correct syntax

   func updateValue(val *int) {} is correct syntax, with * before the type. func updateValue(val int) {} declares a value parameter. func updateValue(&val int) {} and func updateValue(*val int) {} have invalid syntax.

3. Final Answer:

   func updateValue(val *int) {} -> Option A

4. Quick Check:

   Pointer parameter syntax = *type [OK]

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

Common Mistakes:

• Putting * before variable name instead of type
• Using & in parameter declaration
• Confusing pointer syntax with value parameter

3. What is the output of this Go program?

package main
import "fmt"
func changeValue(val int) {
    val = 10
}
func main() {
    x := 5
    changeValue(x)
    fmt.Println(x)
}

medium

A. 10

B. 5

C. 0

D. Compilation error

Solution

1. Step 1: Analyze function parameter passing

   The function changeValue receives x by value, so val is a copy of x.

2. Step 2: Check if original variable changes

   Inside changeValue, val is set to 10, but this does not affect x in main.

3. Final Answer:

   5 -> Option B

4. Quick Check:

   Passing by value means original x stays 5 [OK]

Hint: Passing by value means original variable unchanged [OK]

Common Mistakes:

• Assuming x changes to 10 after function call
• Confusing pointer and value passing
• Expecting compilation error due to assignment

4. Identify the error in this Go code snippet:

func increment(num *int) {
    num = num + 1
}

func main() {
    x := 5
    increment(&x)
    fmt.Println(x)
}

medium

A. Missing & operator when calling increment.

B. x is not initialized before use.

C. Function parameter should not be a pointer.

D. Cannot add 1 to a pointer directly; need to dereference first.

Solution

1. Step 1: Understand pointer arithmetic in Go

   Go does not support arithmetic on pointers like 'num + 1'. You must dereference the pointer to change the value.

2. Step 2: Correct way to increment value pointed by pointer

   Use '*num = *num + 1' to increment the value at the pointer address.

3. Final Answer:

   Cannot add 1 to a pointer directly; need to dereference first. -> Option D

4. Quick Check:

   Pointer arithmetic invalid; dereference to modify [OK]

Hint: Use * to access pointer value before arithmetic [OK]

Common Mistakes:

• Trying to add directly to pointer variable
• Forgetting to dereference pointer
• Confusing pointer and value types

5. You want to write a function in Go that doubles the value of an integer variable passed to it. Which approach correctly modifies the original variable?

hard

A. Return the doubled value and assign it outside the function.

B. Pass the integer by value and assign double inside the function.

C. Pass a pointer to the integer and modify the value via dereferencing.

D. Use a global variable instead of passing parameters.

Solution

1. Step 1: Understand how to modify original variable

   Passing by value copies the variable, so changes inside function won't affect original.

2. Step 2: Use pointer to modify original variable

   Passing a pointer allows the function to access and change the original variable's value via dereferencing.

3. Step 3: Consider alternatives

   Returning doubled value requires assignment outside; using global variables is bad practice.

4. Final Answer:

   Pass a pointer to the integer and modify the value via dereferencing. -> Option C

5. Quick Check:

   Pointer passing modifies original variable [OK]

Hint: Use pointer to change original variable inside function [OK]

Common Mistakes:

• Assuming passing by value changes original variable
• Relying on return value without assignment
• Using global variables unnecessarily