Error storing address in pointer?

Question

PHP

The code:

int Europe_CountryCount=44;
    int** nextWorldPart = new int*[6];
    nextWorldPart[0]=&Europe_CountryCount;

Mistake:

error: 'nextWorldPart' does not name a type
nextWorldPart[0]=&Europe_CountryCount;

The above code is in .h file.
Cleaning and rebuilding the project didn't help.

Answer 1

$CONNECT = mysqi_connect('localhost' ,'u8310614_*****', '******' ,'u8310614_****');

missing ;at the end of the line.
well, mysqli_connectinsteadmysqi_connect

Answer 2

See what kind of error is happening. mysqli allows you to do this:

$CONNECT = mysqi_connect('localhost' ,'u8310614_*****', '******' ,'u8310614_****');

if (!$CONNECT)
    echo "Код ошибки errno: " . mysqli_connect_errno() . PHP_EOL;
    echo "Текст ошибки error: " . mysqli_connect_error() . PHP_EOL;
    exit;
}

Answer 3

nextWorldPart[0]=&Europe_CountryCount; It is neither a declaration nor a definition. This is an operator call that can only be executed in a function.

- answer by AR Hovsepyan
https://ru.stackoverflow.com/q/1100821/203448