* is both a binary and a unary operator in C and it means different things in different context. Based on the code you have provided: *leaf = (struct node*) malloc( sizeof( struct node ) ); Here the void * (void pointer) that malloc returns is being casted to a pointer to struct node, I … Read more
Parse error: syntax error, unexpected ‘else’ (T_ELSE) in C:\xampp3\htdocs\wb_html\login.php on line 8 [duplicate] solved Parse error: syntax error, unexpected ‘else’ (T_ELSE) in C:\xampp3\htdocs\wb_html\login.php on line 8 [duplicate]
String literals can be concatenated like that, but string values cannot. Further, it seems that you want the output of gcommand to end up in the buffer output. It is not possible to do that with the system function. Assuming you are going to be executing in a POSIX-style shell where > is the shell … Read more
I fixed your code using Swift 3 syntax. class SignUp: UIViewController { @IBOutlet weak var buttonNameTxt: UITextField! @IBOutlet weak var buttonEmailTxt: UITextField! @IBOutlet weak var buttonPwdTxt: UITextField! override func viewDidLoad() { super.viewDidLoad() } @IBAction func buttonSignIn(_ sender: UIButton) { let usermainname = buttonNameTxt.text! let username = buttonEmailTxt.text! let password = buttonPwdTxt.text! let myURL = URL(string: … Read more
I figured it out; it is because the arguments of the javascript function that get written on the updated page are strings and so need qoutes around them (one of them had spaces and thats why i didn’t get a line number it messes up the parser) solved JavaScript Syntax error
No, it’s a syntax error. It can’t be part of a multiplication since there cannot be an identifier with the name of a keyword. The only way to make this a valid part of a C program is as part of a string literal as part of a multibyte character constant as part of a … Read more
Each of the parameters are being used to find particular X and Y values. O changes from 0 to pi in steps of pi/8 while Vo, t and g remain unchanged. The t variable is simply an array from 0 to 5 in steps of 1 and so there are 6 time points defined all … Read more
Of course you can: #define S str##ing #define S_INCLUDE <S> #include <iostream> #include S_INCLUDE int main() { std::S str = “Hello world!”; std::cout << str << std::endl; } 2 solved Is there any way to use string in c++ without the word “string” appear in the program? [closed]
Introduction The Swift programming language is a powerful and versatile language that allows developers to create robust and efficient applications. One of the features of Swift is the ability to create enumerations, or enums, which are used to define a set of related values. However, when declaring an enum case, it is important to be … Read more
Swift enumeration cases are defined as case someName, not case .someName. This is an easy syntax error when declaring a new enum’s cases, as in most other situations you will be typing .someName via dot syntax. But when first declaring that enum case, it’s case someName without the period. enum SomeEnum { case one case … Read more
Multiple issues here: assignment from input into a variable cannot be checked in while-loop, you should split it into assignment and check. else cannot contain a condition you also had a bug in printing the results – you printed num1 twice The indentations are meaningful in Python – please make sure to post it properly … Read more
While making a function virtual in c++ where do I have to write virtual keyword? In the function declaration, before the function name, and after any attribute specifiers, along with the other specifiers (including the type specifier for the function’s return type). The general syntax for a declaration is simple-declaration: decl-specifier-seq<opt> init-declarator-list<opt> ; attribute-specifier-seq decl-specifier-seq<opt> … Read more
It looks like you’re referring to the wrong scope for your operator[]. Perhaps you meant: CScreen::Proxy CScreen::operator[] (int index) const { return Proxy ( m_playground, index, m_y ); } 2 solved Proxy class and correct syntax c++ [closed]
Your first code reads as $a = false; if ($a) { echo ‘A’; } if (false) { echo ‘B’; } else { echo ‘C’; } When you do not use braces, only the next statement (statement, not line) gets executed as part of that control structure. It is not ambiguous as this behaviour is clearly … Read more
Fella…that was hard to believe…. if ($currentpage == ‘/url-1’ || $currentpage == ‘/url-2’) { solved PHP if a = 1 or 2 then do something