无法从另一个函数C ++类访问成员

Question

所以基本上我的代码如下：

void SimpleMQTT::begin() {
    Serial.println("setserver");
    WiFiClient espClient;
    PubSubClient client(espClient);
    client.setServer(this->serverAddress, this->port);
    while(!client.connected()) {
        if (client.connect(this->deviceName)) {
            Serial.println("connected to mqtt");
            client.publish("connecting", "connected");
        } else {
            Serial.println("failed to connect");
        }
    }
    this->client = & client;
    //WORKING
    this->client->publish("connecting","conn2");
}

void SimpleMQTT::broadcast(char* channelName, char* message) {
    Serial.println("broadcasting");
    Serial.println(this->deviceName);
    //NOT WORKING
    this->client->connected();
    Serial.println("after broadcast");
}

可以通过功能begin()访问成员，但是不能使用功能broadcast()访问同一成员。 Arduino正在向Serial发送异常（请参见屏幕截图）。

Answer 1

您似乎有一个悬而未决的指针问题。 您在堆栈上创建客户端对象：

PubSubClient client(espClient);

然后引用它：

this->client = & client;

但是，一旦函数SimpleMQTT::begin()退出，client就会与函数的其余堆栈一起删除。

您应该改为在堆上创建客户端对象。 将代码更改为：

void SimpleMQTT::begin() {
    Serial.println("setserver");
    WiFiClient* espClient = new WiFiClient();
    PubSubClient* client = new PubSubClient(espClient); // Allocate an object on the heap
    client.setServer(this->serverAddress, this->port);
    while(!client->connected()) {
        if (client->connect(this->deviceName)) {
            Serial.println("connected to mqtt");
            client->publish("connecting", "connected");
        } else {
            Serial.println("failed to connect");
        }
    }
    this->client = client;
    this->client->publish("connecting","conn2");
}

espClient也应该是您班上的成员。

这里是对danling pointers的更详尽的解释。