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I am getting a warning saying I have reached end of non-void function


By : Vivek Radhakrishnan
Date : July 31 2020, 06:00 AM
hope this fix your issue Others have pointed out your error and offered alternative ways to code this. Let me offer another. It's an early-return method that keeps the code flatter, shorter, and easier to understand:
code :
bool is_leap(int y) {
    if (y % 400 == 0) return true;
    if (y % 100 == 0) return false;
    return y % 4 == 0;
}


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Control reached end of Non Void Function


By : user3783758
Date : March 29 2020, 07:55 AM
seems to work fine Wihtout knowing the exact error message, I am not 100% sure what is the problem, but I guess it is the following:
Your function looks like this (simplified):
code :
int calc(int x) {
    if(someCondition){   
        if(otherCondition){}
        calc(x+1);
    } else {
        return mul;
    }
}
return calc(x+1);
calc(x+1);

'file_get_contents' function from multiple URLs and redirection limit reached warning


By : user2227380
Date : March 29 2020, 07:55 AM
I hope this helps . I would suggest using cURL for fetching remote data. You could do this:
code :
$urls = [
    "https://www.url1.com",
    "https://www.url2.com",
    "https://www.url3.com",
    "https://www.url4.com",
    "https://www.url5.com"
  ];
$decoded = array_map("loadJSON", $urls);

if (is_array($decoded[0])) {
  foreach ($decoded[0] as $key => $value) {
    if (is_array($value) && isset($value['price'])) {
      $price = $value['price'];
      echo '<span><b>' . $price . '</b><span>';
    }
  }
}

/**
 * Downloads a JSON file from a URL and returns its decoded content
 */
function loadJSON($url) {
  $ch = curl_init();
  curl_setopt($ch, CURLOPT_URL, $url);
  curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
  curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, 0); // If your server does not have SSL
  curl_setopt($ch, CURLOPT_FOLLOWLOCATION, 1); // Follow redirections
  curl_setopt($ch, CURLOPT_MAXREDIRS, 10); // 10 max redirections
  $content = curl_exec($ch);
  curl_close($ch);
  $res = json_decode($content, true);
  return $res;
}

static inline void function has warning "control reaches end of non-void function"


By : the nightfly
Date : March 29 2020, 07:55 AM
I hope this helps you . I currently have the following function in C: , void * is not void
Change your first line to
code :
static inline void cmyk_to_rgb(int *dest, int *cmyk)

warning: ‘return’ with a value, in function returning void return next; ex19.c:95:10: error: void value not ignored as i


By : Jeanielmj
Date : March 29 2020, 07:55 AM
wish helps you I am going through the Learn Code the hard way book and stuck on exercise 19 http://c.learncodethehardway.org/book/ex19.html. I got desperate, copied and pasted the code, yet still I am getting errors: , This line:

I have a warning and error with void value; warning:in function returning void, error: void value not ignored


By : G Turner
Date : March 29 2020, 07:55 AM
will be helpful for those in need void print_month(...) means this function will return no values. But you are returning a value at the end - return first_day;
Change void print_month(...) to int print_month(...) to make it return an integer value.
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