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Responsible Gambling
Oct 15, 2017
LOPSA is an independent, not-for-profit organisation created by the Australian . The peak body for the Australian poker community, LOPSA provides a social and commercial network that connects.Cinematic Promotions

Cinematic promotions are the cornerstone of a successful sponsorship and are a key element in the creation of a brand-building program. They are also a highly leveraged tool for event marketing.

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Examples of Cinematic Promotions

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Contact us today and find out more about how Cinematic Promotions can help your brand.Q:

Defining function in C

I’m trying to understand how to declare a function in C.
Here is my code:
int add(int a, int b)
return a + b;

Can someone tell me why I can’t call the add function like this:
printf(“The result is: %d”, add(3, 4));

I read that you can’t use the * or [] for invoking functions so I try this.


You cannot overload a function for different arguments.
All functions with the same name are of the same type
Your function declaration can be rewritten
int add(int a, int b) { return a + b; }


You cannot use the [] or * in a declaration of the function name, as in:
int add(int a, int b);
int add(int a, int b)
return a + b;

What is going to happen is that the declaration of the function has a declaration


In your case you have a error in the variable $toupload
$toupload = ‘../cache/’;

Should be
$toupload = ‘../cache/’. $id;

This is because you are concatenating a integer to a string in your $toupload var.
$_SESSION[‘userid’] = $id;

$insert = mysqli_query($con, “INSERT INTO public_to_private(userid, private_key, public_key, platform, notes, date) VALUES ($userid, ‘$toupload’, ‘$toupload’, ‘$platform’, ‘$notes’, now()) ON DUPLICATE KEY UPDATE public_key = ‘$toupload’ WHERE userid = ‘$userid'”) or die(mysqli_error($con));

if ($insert) {
echo ‘OK’;
} else {
echo ‘Something went wrong, check connection’;

$userid = mysqli_insert_id($con);

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