I believe that clean and elegant retry code is notoriously difficult, so I'm wondering how the following code could be improved. The retryGet function should have the same interface and contract as http.Get, except for the number of attempts argument, of course.

func retryGet(url string, maxAttempts int) (*http.Response, error) {
    attempts := 0

    if maxAttempts < 1 {
        return nil, errors.New("maxAttempts must be at least 1")

    for {
        response, err := http.Get(url)
        if err == nil && response.StatusCode == http.StatusOK {
            return response, nil

        delay, retry := shouldRetry(maxAttempts, attempts, response)
        if !retry {
            return response, err

        if err == nil {
            defer response.Body.Close()


func shouldRetry(maxAttempts, attempts int, response *http.Response) (time.Duration, bool) {
    if attempts >= maxAttempts {
        return time.Duration(0), false

    delay := time.Duration(attempts) * time.Second

    if response != nil && response.Header.Get("Retry-After") != "" {
        after, err := strconv.ParseInt(response.Header.Get("Retry-After"), 10, 64)
        if err != nil && after > 0 {
            delay = time.Duration(after) * time.Second

    return delay, true

I would recommend you to actually strictly match the http.Get signature using a struct:

type Getter func(url string) (*http.Response, error)

type RetryingClient struct {
    Getter Getter
    MaxAttempts int

func (rc RetryingClient) Get(url string) (*http.Response, error) {

And instead of directly calling http.Get, the embedded Getter interface gives you much more flexibility:

  • for testing
  • for using a proxy

Why do you defer the response.Body.Close() ? Since it won't be use, you can close it immediately.

Using the interface, you could make a function embedded in the struct to compute a default delay instead of hard-coded time.Duration(attempts) * time.Second. For instance: defaultDelay(attempt int) time.Duration


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.