Question

这是一个不同的问题，因为它与我的代码没有直接关系，而是与another question中导致问题的代码直接相关。

在挖掘与其他问题相关的代码时，我发现了一个我不理解的奇怪的return语句。在继续我的问题之前，请查看laravel/framework代码中的这两个代码段。

特质\Illuminate\Foundation\Auth\AuthenticatesUsers：

/**
 * Handle a login request to the application.
 *
 * @param  \Illuminate\Http\Request  $request
 * @return \Illuminate\Http\RedirectResponse|\Illuminate\Http\Response|\Illuminate\Http\JsonResponse
 *
 * @throws \Illuminate\Validation\ValidationException
 */
public function login(Request $request)
{
    $this->validateLogin($request);

    // If the class is using the ThrottlesLogins trait, we can automatically throttle
    // the login attempts for this application. We'll key this by the username and
    // the IP address of the client making these requests into this application.
    if ($this->hasTooManyLoginAttempts($request)) {
        $this->fireLockoutEvent($request);

        return $this->sendLockoutResponse($request);
    }

    if ($this->attemptLogin($request)) {
        return $this->sendLoginResponse($request);
    }

    // If the login attempt was unsuccessful we will increment the number of attempts
    // to login and redirect the user back to the login form. Of course, when this
    // user surpasses their maximum number of attempts they will get locked out.
    $this->incrementLoginAttempts($request);

    return $this->sendFailedLoginResponse($request);
}

特质\Illuminate\Foundation\Auth\ThrottlesLogins：

/**
 * Redirect the user after determining they are locked out.
 *
 * @param  \Illuminate\Http\Request  $request
 * @return void
 * @throws \Illuminate\Validation\ValidationException
 */
protected function sendLockoutResponse(Request $request)
{
    $seconds = $this->limiter()->availableIn(
        $this->throttleKey($request)
    );
    throw ValidationException::withMessages([
        $this->username() => [Lang::get('auth.throttle', ['seconds' => $seconds])],
    ])->status(429);
}

在AuthenticatesUsers特征中，检查是否为给定用户进行了太多的认证尝试。如果是，则触发事件并返回锁定响应。到目前为止一切都很好。

我不理解的是return前面的$this->sendLockoutResponse($request)语句。给定方法确实总是抛出一个异常并且什么都不返回（好吧，它会返回void，但它没有，因为它总是throws）。

那么这里return语句的目的是什么？这是否为读者提示此时login()被取消或者这是一些特殊的语法我以前从未听说过？

如果永远不会达到return语句，为什么返回void？

0 个答案: