golang如何去除 context 的 deadline

func main() {
	ctx := context.WithValue(context.Background(), "trace", "123")
	ctx, cancel := context.WithTimeout(ctx, time.Second)
	defer cancel()

	go func(ctx context.Context) {
		newCtx := context.WithValue(context.Background(), "trace", ctx.Value("trace"))
		// do something with newCtx
	}(ctx)

	fmt.Println("main finished")
}

type Context interface {
	Deadline() (deadline time.Time, ok bool)
	Done() <-chan struct{}
	Err() error
	Value(key any) any
}

// contextWithoutDeadline 伪造了一个没有 deadline 的 context
type contextWithoutDeadline struct {
	ctx context.Context
}

func (*contextWithoutDeadline) Deadline() (time.Time, bool) {
	return time.Time{}, false
}

func (*contextWithoutDeadline) Done() <-chan struct{} {
	return nil
}

func (*contextWithoutDeadline) Err() error {
	return nil
}

func (l *contextWithoutDeadline) Value(key interface{}) interface{} {
	return l.ctx.Value(key)
}

// DetachDeadline 从 context 剥离 deadline
func DetachDeadline(ctx context.Context) context.Context {
	return &contextWithoutDeadline{ctx: ctx}
}

// SetNewTimeout 设置新的超时时间
func SetNewTimeout(ctx context.Context, timeout time.Duration) (context.Context, context.CancelFunc) {
	return context.WithTimeout(DetachDeadline(ctx), timeout)
}

func main() {
	ctx, _ := context.WithTimeout(context.Background(), time.Duration(10)*time.Second)
	fmt.Println(ctx.Deadline())

	newCtx := contextwarp.DetachDeadline(ctx)
	fmt.Println(newCtx.Deadline())

	newCtx2, _ := contextwarp.SetNewTimeout(ctx, time.Duration(1)*time.Second)
	fmt.Println(newCtx2.Deadline())
}