Why Do Go's Struct Setter Methods Behave Differently When Using Value vs. Pointer Receivers?

Understanding Setter Methods for Struct Types

Struct types in Go provide a convenient way to group related data, and setter methods allow us to modify their properties. However, certain scenarios can lead to unexpected behavior.

Problem Description:

Consider a struct T with a field Val and two setter functions: SetVal and SetVal2. Using SetVal does not modify the original struct, while SetVal2 does. Understanding this discrepancy is crucial.

Underlying Mechanism:

When passing a struct to a function, two approaches are possible:

• Passing by Value: Creates a copy of the struct. Any modifications made within the function affect only the copy.
• Passing by Reference (Pointer): Provides a pointer to the original struct, allowing modifications to persist.

Reasoning:

SetVal takes a struct as a value parameter. Therefore, a copy of the struct is created within the function, and any changes to t (the copy) do not impact the original v.

Resolving the Issue:

Use the pointer receiver approach in SetVal2 to ensure modifications are reflected in the original struct:

func (t *T) SetVal(s string) {
    t.Val = s
}

Verification:

Adding print statements to demonstrate the difference:

type T struct { Val string }

func (t T) SetVal(s string) {
    fmt.Printf("Address of copy is %p\n", &t)
}

func (t *T) SetVal2(s string) {
    fmt.Printf("Pointer argument is %p\n", t)
}

func main() {
    v := T{"abc"}
    fmt.Printf("Address of v is %p\n", &v)
    v.SetVal("pdq")
    v.SetVal2("xyz")
}

This program outputs:

Address of v is 0xf8400cfc00
Address of copy is 0xf8400cfcd0
Pointer argument is 0xf8400cfc00

The addresses of v and the pointer in SetVal2 are equal, confirming the usage of the original struct, while SetVal works on a copy.

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