Pointy’s answer has good information, but it’s not the correct answer for this question. The behavior described by the OP is part of a bug that was first reported in March 2010, patched for Webkit in August 2012, but as of this writing is not yet integrated into Google Chrome. The behavior hinges upon whether … Read more
If a is an int variable, is it always true that *&a == a ? It should be, but I can’t find a definitive statement to that effect. If p is an int* variable, is it always true that p == &*p ? Yes – in that case, neither the & nor * operators are … Read more
You would do so by creating the variable ($value as I’ve done below) and then concatenating the $url string using . $value=”90″; $url = “http://junna.com/mamba” . $value . “nina”; 6 solved how to put a variable inside a variable value PHP [closed]
What happens to b? Nothing happens to b. When you do a = b; you’re copying the value stored in b and putting it in a. (You’re not making a an alias of b.) When you then do a += 1; you’re changing the value stored in a (and the value stored in b remains … Read more
Consider Apollo’s comment and google it. This is not only a java principle but a generall programming principle. You’re creating a new Answer Object. Let’s look at this example: public class Answer{ // The class private String answer; // internal variable only visible for the class public Answer(String answer){ // This is a constructor this.answer … Read more
The order in which the arguments of printf are evaluated is implementation specific. Output change from machine to machine. solved Assignment aand Precedence of operators in c [duplicate]
Yes, the output is correct: // This line increments the variable then prints its value. Console.WriteLine(“{0}”, ++ Cash); // This prints the value of the (incremented variable) Console.WriteLine(“{0}”, Cash); // The prints the value of the variable *then* increments its value Console.WriteLine(“{0}”, Cash ++); 1 solved C# help declaring variable i initialize it to
There are two key elements you should know about: Use a variable to store the object and then pass it as a parameter to a method. Part 1: a variable After creating an object, you should store it in a variable. Otherwise it just exists in memory and will be garbage collected and thus be … Read more
If you like to create a piece of code having that many different variables with a different name, I suggest you start writing your own code generator. The code below should be a good start. #include <sstream> #include <iostream> #include <string> #include <vector> namespace { template <typename Function> void bitExecutor(int nrOfBits, std::vector<bool> &bits, const Function … Read more
You can’t access $dine in your second function because it’s nowhere defined. The $dine from your first function is only a local variable. I would suggest this solution which uses the fact that calculate_dinein_price() also returns the value of $dine: public function calculate_dinein_total(){ $total_dine = 0.00; $total_dine = $total_dine + $this->calculate_dinein_price(); return $total_dine; } 8 … Read more