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How could an integer variable store a reference to integer?
No, it is fine either way. The return value reference is not even necessary in this special case because you are not trying to modify the return value "on the fly" or the 'a' later, like when you use arithmetic operator overloads for that purpose, for instance

Categories : C++

Why is a .mp3 passed as an integer?
This is just like any other reference that you access through R (e.g. R.string.x, R.dimen.x). You can access your files in raw through R.raw.x (so in this case, R.raw.sound01). The integer is just a guaranteed unique identifier for that particular resource.

Categories : Android

Ensure variable is passed as an integer
I would consider that a buggy API, since Perl considers signed integer (IV) four, unsigned integer (UV) four, floating point number (NV) four, 8-bit string (PV,UTF8=0) four and 72-bit string (PV,UTF8=1) four as the same value. (And that's not even counting overloaded objects and magical variables!) You can convert the representation of the value to one of the numerical formats (IV, UV or NV) using an operation that takes a number and returns a number: 0+$code If you want an IV specifically, you might have to access the perlapi.

Categories : Json

Cant Get the integer value of the response passed from server in an ajax call?
Use the PrintWriter#print(int i) method. Prints an integer Parameters: i - The int to be printed PrintWriter#write(int c) method says that Writes a single character. Parameters: c - int specifying a character to be written. Hence you are seeing the corresponding ASCII character on using write() method.

Categories : Java

PHP: Only variables can be passed by reference
You can't pass a constant of 1, a fix is to set it to a variable as so. Change: $password = str_replace($key, $value, $password, 1); to: $var = 1 $password = str_replace($key, $value, $password, $var); UPDATE: Changed to declare variable outside of the method call from feedback in comments.

Categories : PHP

Parameters passed by reference can not set value
You are assigning the textbox value to the ck field in the form, so get it from there: string _check_code = "0000"; var checkcodeForm = new CheckcodeForm(_check_code,checkcodePicUrl).ShowDialog(); MessageBox.Show(checkcodeForm.ck); BTW, you are passing the parameter by value, not by reference. Check Passing Parameters (C# Programming Guide)

Categories : C#

Is context passed by reference
Yes, it is passed by reference. ApplicationContext is a class and every class variable in C# is passed by reference into a method. Calling context.SaveChanges makes it clearer that a whole unit of work is committed that could span several repositories, not only one. For example you could create a second repository var roleRepo=new Repository<Role>(context); and do some operations like adding a new Role. context.SaveChanges will commit all changes and insert the added user and the added role in a single transaction. Using userRepo._context.SaveChanges() (which is technically possible with a public _context and would do exactly the same) might lead to the wrong idea that only changes done in the user repository are saved. But in fact all changes done with any repository that got the

Categories : Entity Framework

Garbage value when passed float values to the function accepting integer parameters
You need to provide a declaration for func before calling it int func(int a,int b); int main() { } int func(int a,int b) { // implementation } Your code presumably compiled with a warning something like warning: implicit declaration of function 'func'. If you'd fixed this warning, you'd also have fixed the error. It'd be good practice to have the compiler warn you about as many potential problems as possible. You may also want to guard against the possibility of missing warnings by treating them as errors. You can do this by adding -Wall -Werror to your build command for gcc or /W4 /Wx for MSVC. Note that the correct value is returned if you pass ints into func. At a guess, this may be because the calling code doesn't know that it needs to cast the arguments to int and passes

Categories : C

What causes this list to be passed by reference when called one way, but by value another?
ints(b) call does not enumerate IEnumerable, so code never reaches l.Add(4) line, unlike in AddRange case which enumerates all items to add them to the list. To see it called for b case enumerate result manually: ints(b).ToList(); IEnumerable<T> implemented via functions are not executing body of the function before enumeration starts - the code is actually transformed by compiler into class with states to support true lazy evaluation of enumerable (details can be found in multiple articles, i.e. Iterator Pattern demystified - link provided by Tim Schmelter).

Categories : C#

Result of function can't be passed as reference
When you do this: f(get()); you pass a temporary std::list<int> to f(). A temporary cannot bind to a non-const reference. So you can fix this by passing a const reference, since you do not want to modify the argument. void f(const std::list<int>& v) { // ^^^^^ for (std::list<int>::const_iterator i = v.begin(); i != v.end(); ++i) { // ^^^^^^^^^^^^^^ printf("Hello %d ", *i); } } Note that this requires that you use a const_iterator, since std::list::begin() const and the corresponding end() method return const_iterators. In C++11 you can simplify this to for (auto i = v.begin(); i != v.end(); ++i) ... or even for (const auto& i : v) std::cout << i << " ";

Categories : C++

Object not being passed by reference (CoffeeScript)
@data is being passed by reference but you're throwing that reference way when you pull out the Data.foo value here: @foo = Data.foo Your Data is the same as your @data but @foo is just the value of Data.foo. You'd need to hold onto Data rather than Data.foo: class Mutator constructor: (Data) -> @data = Data and then look at @mutator.data.foo. You could probably use defineProperty to define a getter for foo and thus hide the data part and make @mutator.foo work. Have a look at this answer for some notes on how this would work.

Categories : Javascript

In Javascript, how is this argument passed by value and not by reference?
You get "Nicholas" precisely because JavaScript is never "by reference". If it was, you'd be able to update the person variable from any location. That's not the case, so the new Object() in the function doesn't mutate the outer person variable. But it's also not the case that variables refer to the objects themselves, but rather variables hold a special type of reference that let you update the object without directly accessing the memory. That's why although JavaScript is always "by value", you never get a full copy of an object. You're merely getting a copy of that special reference. As a result, you can manipulate the original object passed via the copy of the reference, but you can't actually replace it via the reference.

Categories : Javascript

How do I reference the object passed to my class when I instantiate?
The problem here is that you have header_path declared as a variable for the init function. It's scope is local to the init function. What you need is to associate header_path as a variable for the class instance. Class SECHeader(object): def __init__(self,header_path): self.header_path = header_path # Instantiate a variable for class object self.header = open(header_path).read() def parsed_name(self): return self.header_path.split('-')[-1] # Call the associated variable Another way is to actually call the variable that you gave as an argument to SECHeader in parsed_name. This variable name would be in the class namespace. for banana in header_paths: header = SECHeader(banana) print header.parsed_name() Class SECHeader(object): def __init_

Categories : Python

php Strict Standards: Only variables should be passed by reference in "use"
Here http://php.net/manual/en/language.references.pass.php it says that "There is no reference sign on a function call - only on function definitions." Try removing the '&' from your function call code there and see if that gets rid of the message. ---Edit--- Looking at this thread here "Strict Standards: Only variables should be passed by reference" error you could try saving your callback function into a variable before passing it to the array walk function: $consoldatedCities=array(); $callbackFcn= function($cityName,$cityId) use(&$consoldatedCities) { $consoldatedCities[$cityId] = $cityName; }; array_walk_recursive($StateCityHash, $callbackFcn);

Categories : Misc

Can arguments be passed just by reference to function in javascript
When you add an event listener the function you add gets passed an event object depending on what type of event it is (click, scroll, etc). http://www.w3schools.com/jsref/dom_obj_event.asp window.addEventListener("click", function(event) { alert(event.pageX + ", " + event.pageY); }, false); // same as window.addEventListener("click", function(e) { alert(e.pageX + ", " + e.pageY); }, false); The parameter name can be anything it still represents the event object. EDIT List all of the properties and functions of an event object window.addEventListener("click", function(event) { var all = ""; for(var prop in event) { all += prop + " : " + event[prop] + " "; } alert(all); }, false);

Categories : Javascript

Strict Standards: Only variables should be passed by reference in..
BTW your codes works for me. And i dont know about it works but try this one function AFM_fileExt($filename) { $arr = explode('.', $filename); return strtolower(end($arr)); }

Categories : PHP

Array.map+try/catch passed reference assignment delay
The map function is synchronous, but inside the callback function that you passed to the map function, you are performing asynchronous operations. reader.readAsText( file ); is asynchronous. That means the map function returns before any data will get appended to the parsed array. A simple fix would be to keep track of how many files have been processed by adding counters in the onload and onerror callbacks. When the number of files that have been processed reaches flist.length, you would call a function that would do the work with parsed.

Categories : Javascript

Array to string conversion and Only variables should be passed by reference
only variables should be passed by reference You are passing the result of a function call as an argument. You aren't passing a variable. $vars = get_defined_vars(); echo var_name($a,$vars); Also, unless you're intentionally modifying one of the variables you shouldn't be passing it as a reference. That way any changes made are local to the function.

Categories : PHP

In groovy when comparing this and a reference passed in a method: same instance, different metaclass
You can implement methodMissing in the class as below to answer your last question: class Greeter { def sayHello() { //this.metaClass.greeting = { System.out.println "Hello!" } greeting() goodNight() } def methodMissing(String name, args){ if(name == 'greeting'){ println "Hello!" } else println "Good Night" } } new Greeter().sayHello() Also note that == in groovy actually means equals() (that is value comparison) if you want to compare identity then is() can be used like a.is(b) //Corresponds to == in Java a == b //Corresponds to equals() in Java UPDATE Can use metaClass as below Greeter.metaClass.greeting = { println "Hello"} def greet = new Greeter() //or //greet.metaClass.greeting = { println "Hel

Categories : Groovy

Strict standards: Only variables should be passed by reference in get_id.php on line 27
there are a lot of similiar questions here. end expects its parameter to be a passed by reference, and only variables can be passed by reference (not the return value of another function, like preg_split in your case) easiest solution should be to split the calls to separate lines, e.g. $splitted = preg_split('/<.*>| /', $output, -1, PREG_SPLIT_OFFSET_CAPTURE); $last_item = end($splitted); $very_last_item = end($last_item); or, if you have the time and stuff, rework this function so you don't have to end() calls (sorry, I kinda can't get into what you want to accomplish here)

Categories : PHP

Using a passed by reference string array to runtime loaded dll function
Ok, it seems I found the answer to all such problems... Namely, I'm trying to return a C++ class (std::string) in a "extern "C"" function. It was just a matter of making it return a standart const char* that everything started to run just fine. Thanks for the contributors, Momergil

Categories : C++

Strict standards: only variables should be passed by reference - why does this still work and how can I correct it?
Can't answer your first question, but here is how you pass by reference: //check if email already taken if ($stmtreg = $mysqli->prepare("SELECT user_email FROM users WHERE user_email = ?")) { $email = strtolower($_POST['srEmail']); $stmtreg->bind_param("s", &$email); $stmtreg->execute(); $stmtreg->store_result(); $num_rows = $stmtreg->num_rows(); $stmtreg->bind_result($email); $stmtreg->fetch(); $stmtreg->close(); } In order to pass a variable by reference, you need to put an & before the variable name (eg &$email)

Categories : PHP

Strict Standards: Only variables should be passed by reference - Registration Form
$query->bindParam(':name', $user->get("name")); //ERROR HERE You must put the data into a variable, before passing it to the function, requiring a reference as a parameter. Like this: $variable = $user->get("name"); $query->bindParam(':name', $variable); //NO ERROR HERE Here is the explanation, what can be passed by reference: http://www.php.net/manual/en/language.references.pass.php

Categories : PHP

MFC WebBrowser.Navigate error "A null reference pointer was passed to the stub."
Try using BSTR string instead of the literal _T("about:blank") and pass an empty variant instead of NULL: COleVariant vUrl(_T("about:blank")); COleVariant vEmpty; m_browser->Navigate(V_BSTR(&vUrl), &vEmpty, &vEmpty, &vEmpty, &vEmpty); That matches the signature of IWebBrowser2::Navigate method: HRESULT Navigate( BSTR url, VARIANT *Flags, VARIANT *TargetFrameName, VARIANT *PostData, VARIANT *Headers );

Categories : C++

Most RAM efficient way to reference integer constants in C
For both #define and enum the compiler will be able to code the integer constant into the generated instructions. It may also be able to do it with const int but that depends on the optimization capabilities of your compiler. The extern will almost always need to reference an address, which means the value will be somewhere else taking up memory. Yes, the constant will be stored more than once, but that's preferable to a pointer to the constant being stored more than once.

Categories : C

Reference values: string or integer?
From the sounds of your description, these are values that currently happen to be represented by a series of digits; they are not actually numbers in themselves. This, incidentally, is just like my phone number: it is not a single number, it is a set of digits. And, like my phone number, I would suggest storing it as a string. Leading zeros don't appear to be an issue here but considering you are treating them as strings, you may as well store them as such and give yourself the future flexibility.

Categories : C#

Strict Standards: Only variables should be passed by reference in wordpress/wp-includes/class-oembed.php on line 116
You should not use assignments (=) inside your ternary expression as you will run into problems with operator precedence. You could write it like: $content_oembed = (wp_oembed_get($url) !==false) ? ('<div class="audio" style="width:100%; overflow:hidden;">' . wp_oembed_get($url).'</div>') : ('<audio src="'.$url.'" preload="none" type="audio/mpeg"></audio>');

Categories : PHP

Java: How to print the reference of an Integer Object?
I don't believe you can get the actual memory location of an object in Java (for printing purposes or otherwise). Even when you do System.out.println(new Object()), what you see isn't a "reference" location, it's a string containing the object's hash code.

Categories : Java

static lib call error: Exception: "A null reference pointer was passed to the stub." (using winsock2 Win32)
I was getting this exception "0x000006F4: A null reference pointer was passed to the stub." Turns out disabling my 3rd party firewall stopped the exception being thrown. Perhaps the firewall is intercepting the request and messing something up. Might be worth a try for you :)

Categories : C++

Excel: reference to range in other sheet with integer input
Is the sheet name a given? Assuming yes then try using OFFSET like this: =OFFSET(Sheet1!$A$1,Z1-1,Z2-1,Z3,Z4) Where Z1 to Z4 contain, 1 (for start row position), 1 (for start column position), 9 ( for height of range) and 3 (for width of range) respectively. First argument is always A1 If you want 3 and 9 to represent end point rather than size then use Z3-Z1+1 and Z4- Z2+1 If you can't make that work then a less ambiguous set of input values would be useful - what range do you expect if inputs are 4, 5, 6 and 7? Note that to "see" the output you need to use the OFFSET function in context, e.g. within a function that expects a range like SUM

Categories : Excel

List.ForEach - Is the object passed into ForEach a reference or a copy?
There isn't an object passed in at all - there's a value of type T. If T is a reference type, that value is a reference. The value is passed by value, just as normal. Now assuming T is some reference type with a member called someOtherList, then your code will indeed affect the object that the value of j refers to - but the value of j is just a reference. The value within the list will be the same reference, so you'll be able to see the effect if you then (say) iterate over the list again. Importantly, if you changed your code to: MyList.ForEach( j => { j = new WhateverYourTypeIs() }); that would not affect the list at all. I suspect you may have a little confusion about how pass-by-value and pass-by-reference work, particularly in the context of C#. I have an article on the t

Categories : C#

Best way of calling a function with different arguments passed by reference when you only need one of these arguments
Acknowledge to Mats Petersson: you can't pass nullptr as the function requires references. If the function has no warranty attached to it, then you should create temporary variables and pass them all in. It would be unsafe to pass the same variable; in case it's used in internal computation. Even if the function works with the same variable passed, there is no guarantee that someone won't change the implementation of that function in a way that causes your code to break. You could be opening yourself up to nasty bugs in the future.

Categories : C++

Models and Data Access Objects, which should be passed and what should be passed
If you wanted to further abstract that into different scenarios involving multiple types of storage, you may want to check out the Factory Pattern. edit: looked at your question again, realized you didnt need half that answer but here's what i removed: The way you design your models could be making things a little confusing. Try designing your models as objects instead of processes. The process of a login would be handled by a controller, whereas a model would be responsible for getting things from a database. This would mean you use your models inside the controllers as a way to complete a task.

Categories : PHP

Matlab fopen fails to open file name passed as string variable but opens when passed as string
Hey sorry the example wasn't helpful. I solved this problem by reverting to an older version of matlab from 2013a to 2012b. Basically matlab 2013a was treating the file handle differently when passed as a variable.

Categories : Matlab

warning: return makes pointer from integer without a cast but returns integer as desired
A void * is a pointer to anything, you need to return an address. void * myfunction() { int * x = malloc(sizeof(int)); *x=5; return x; } That being said you shouldn't need to return a void * for an int, you should return int * or even better just int

Categories : C

Is it possible to convert an integer pointer to the actual integer located at that memory location?
Yes, you can do it using references, like this: int &a(*aPtr); At this point, any changes done to the int stored at aPtr will be reflected in a: *aPtr = 123; cout << a << endl; // 123 is printed *aPtr = 456; cout << a << endl; // 456 is printed Of course in order for this to work you need to ensure that the constant 0x457FB indeed represents a valid address available for reading and writing by your program.

Categories : C++

Algorithm to find smallest integer by swapping a pair of digits in given integer
First count each digit, store it in an array (counts[10]). Going from the left, check the digits (following is the description of the loop): Check that there's a digit in counts which is smaller than it. Pick the smallest one. Exception: 0 is not allowed for the very first digit. If there's one, swap, you're done (exit the loop!). Otherwise decrement the digit in counts, and go for the next digit. For each digit you do O(1) work. So the whole algo is O(n). For swapping you want to use the least significant digits (furthers to the right). You can either store these locations on the initial lookup, or just before swapping you can search for the first matching digit starting from the end.

Categories : Algorithm

Need number returned to be an integer instead of float, error: integer expression expected
Are the decimal places not useful for other things? I wouldn't destroy data just because you don't need it at the moment; you should transform it when you take it out of the database: select trunc(prnct_change) from count_statistics The default behavior of TRUNC() on a number is to remove all decimal places.

Categories : Oracle

Crystal Reports 2011 - Integer to String and back to Integer
You cannot have a single formula that outputs multiple data types. Instead, you can have your formula output a string like you already mentioned: if isnull({Command.DAYS_OUT}) then "Missing" else totext({Command.DAYS_OUT}) Then, to get your font color formatting the way you want, you can use the original integer value: if isnull({Command.DAYS_OUT}) or {Command.DAYS_OUT} < 0 then crRed else if {Command.DAYS_OUT} in 1 to 30 then crYellow else crBlack And then do the same for the font type: if isnull({Command.DAYS_OUT}) then crRegular else if {Command.DAYS_OUT} < 0 or {Command.DAYS_OUT} in 1 to 30 then crBold else crRegular

Categories : String

Find a random integer which creates another integer, when using percentages
I would just select a percentage and work backwards from the answer to arrive at the question: p * x = answer | 0 < p < 100, p = 5k, 100 <= x < 10000 So, pick your percentage: p = (5 * rand(1, 9)) / 100.0; Make sure your 100 <= answer / p < 10000: answer = rand(100, p * 9999); Solve for the 'unknown': x = p / y

Categories : Java



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