Before we discuss about the why calling a method on a nil struct pointer doesn't panic, lets understand about the zero value of program initialization and execution .

The Zero Value

When storage is allocated for a variable, either through a declaration or a call of new, or when a new value is created, either through a composite literal or a call of make, and no explicit initialization is provided, the variable or value is given a default value. Each element of such a variable or value is set to the zero value for its type:
  • false for booleans,
  • 0 for numeric types,
  • "" for strings, and
  • nil for pointers, functions, interfaces, slices, channels, and maps.
This initialization is done recursively, so for instance each element of an array of structs will have its fields zeroed if no value is specified.
Example
package main
import "fmt"
type A struct {
Name string
}
func (a *A) Go() {
fmt.Printf("GO: %v\n", a)
}
func Test(fn func()) {
fn()
}
func main() {
aa := &A{Name: "FOO"}
bb := (*A)(nil)
cc := &A{}
Test(aa.Go)
// The variable bb is nil so calling bb.Go() method is expected to cause a panic
// "runtime error: invalid memory address or nil pointer dereference",
// but the method call succeeds. Why doesn't this panic?
Test(bb.Go)
Test(cc.Go)
}
view raw nil-struct-pointer.go hosted with ❤ by GitHub
GO: &{FOO}
GO: <nil>
GO: &{}
view raw Output hosted with ❤ by GitHub

Final Thought

In the above example variable bb is nil, so calling bb.Go() method is expected to cause a panic runtime error: invalid memory address or nil pointer dereference, but the method call succeeds. Why this doesn't panic? Answer to this question explained in the go specification, this does not panic because: The function to be called by the Expression.Name() syntax is entirely determined by the Type of Expression and not by the particular run-time value of that expression, including nil.