Ошибка invalid conversion from const char to char fpermissive - Ремонт и установка крупной бытовой техники

"i" is not a character, it’s a character array that basically decays to a pointer to the first element.

You almost certainly want 'i'.

Alternatively, you may actually want a lookup based on more than a single character, in which case you should be using "i" but the type in that case is const char * rather than just char, both when defining c and in the base::lookup() method.

However, if that were the case, I’d give serious thought to using the C++ std::string type rather than const char *. It may not be necessary, but using C++ strings may make your life a lot easier, depending on how much you want to manipulate the actual values.

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"i" is not a character, it’s a character array that basically decays to a pointer to the first element.

You almost certainly want 'i'.

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I’m trying to upload the code from here to work with an ESP8266. So I copy/pasted the code to the Arduino IDE and I get this error:

C:Program Files (x86)ArduinolibrariesWiFisrc/WiFi.h:79:9: error: initializing argument 1 of ‘int WiFiClass::begin(char*, const char*)’ [-fpermissive]

int begin(char* ssid, const char *passphrase);
^

exit status 1
invalid conversion from ‘const char*’ to ‘char*’ [-fpermissive]

Can anyone help me with this?

per1234

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asked Jun 21, 2017 at 19:50

This begin() method expects a modifiable character array as its first argument. That’s what you should provide:

char ssid[]           = "YOUR_SSID";      // this is changed
const char* password  = "YOUR_PASSWORD";  // this is fine
[...]
WiFi.begin(ssid, password);

answered Jun 21, 2017 at 20:00

Edgar BonetEdgar Bonet

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The location where you are calling the function begin, has as first parameter a parameter of type const char* instead of char* … remove the const from this argument type.

Probably you have something like

const char* s = ....

...

...begin(s, ...)

Change class s to

char* s = ...

answered Jun 21, 2017 at 19:52

Michel KeijzersMichel Keijzers

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ArduinoJson returns keys and values as const char*.

If you try to put these values into a char*, the compiler will issue an error (or a warning) like the following:

invalid conversion from 'ArduinoJson::Internals::JsonVariantAs<char*>::type {aka const char*}' to 'char*' [-fpermissive]

This happens with any of the following expression:

char* sensor = root["sensor"];
char* sensor = root["sensor"].as<char*>();

// in a function whose return type is char*
return root["sensor"].as<char*>();

The solution is to replace char* by const char*

const char* sensor = root["sensor"];
const char* sensor = root["sensor"].as<char*>();

// change the return type of the function
return root["sensor"].as<char*>();

See also:

API Reference: JsonVariant::as<T>()

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Пример из стандартной библиотеки WiFi.h, в котором в самом начале объявляются две переменные:

// Please input the SSID and password of WiFi
const char* ssid     = " ";           //
const char* password = " ";           //

Я, как честный Буратино, вставляю в кавычки имя своей Wi-Fi сети, а затем — PublickKey, пароль то бишь…

Пускаю на компиляцию — здрасьте-нате:

invalid conversion from 'const char*' to 'char*' [-fpermissive]

Указывают на строку с инициализацией Wi-Fi модуля:

 WiFi.begin(ssid, password);

…и подчеркивают переменную ‘password’.

Но мы же умные! Мы же знаем, что в библиотеке надо посмотреть, как именно объявлены типы данных для этой процедуры! Смотрю:

int WiFiClass::begin(char* ssid, const char *passphrase){}

И вот тут я уже понимаю, что ничего не понимаю! А разве в самом начале, еще перед всеми Void, не было человеческим языком С++ объявлено, что переменная ‘password’ имеет именно тот тип, который нужен процедуре?

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