I am using a non-scalar parameter for my parameter study:
*.server.serviceTime = ${B=exponential(20ms), exponential(35ms)}
However, compared to the other scalar parameters, the B parameter is not shown in the Browse Data section of the results, which I was using until now to export the results of my parameter study:
How can I record the parameter of the exponential distribution (B) that I'm using?
The serviceTime is declared in the .ned as follows:
volatile double serviceTime #unit(s);
If I'm not mistaken you would like to record the mean value of the exponential distribution. Here is an example how the PureAlohaExperiment sample does this:
[Config PureAlohaExperiment]
...
Aloha.numHosts = ${numHosts=10,15,20}
Aloha.host[*].iaTime = exponential(${mean=1,2,3,4,5..9 step 2}s)
i.e. put the interation variable inside the exponential function.
You may put in a NED module a parameter called B. Then, you do the following in the omnetpp.ini:
**.B = ${B=exponential(20ms), exponential(35ms)}
Finally, you record the B NED parameter in the finish() function:
recordScalar("B", par("B"));
There is an option param-record-as-scalar for saving parameter as a scalar. An example of using it:
*.server.serviceTime.param-record-as-scalar = true
However, it doesn't work for volatile parameters (there is an error during finishing simulation). It seems that it is intentionally behaviour to avoid registering "meaningless" random values.
If you really need current random value of volatile parameter, you should record it as a new scalar just after reading it, for example:
double serviceTime = par("serviceTime").doubleValue();
recordScalar("serviceTime 1", serviceTime);
// ... later
serviceTime = par("serviceTime").doubleValue();
recordScalar("serviceTime 2", serviceTime);
Related
I am trying to add two variables together. I believe both contain an integer, but when I draw what is stored within $product->mileage, I receive the following error:
A non well formed numeric value encountered
$oilchange = $request->oilchange_at_kms;
$product = Product::find($request->product_id);
$mileage = $product->mileage; // Error within this variable, but it is an int
$total = $mileage + $oilchange;
How can I test this, or how can I find the problem in my code?
This error usually pops up when you try to add an integer with a string or some type of non numeric field.
You can test this by using the PHP gettype() method:
dump(gettype($product->mileage));
dd(gettype($oilchange));
If it turns out that one of these is a string (possibly from a form response), you can cast it to an int if you are certain that the value will always be an int.
$mileage = (int)$product->mileage;
Not really recommending this, as you should try to resolve the types within the variables first, but it may help you in testing.
Ok, I am starting to get the jist of rvalue references (I think). I have this code snippet that I was writing:
#include <iostream>
using namespace std;
std::string get_string()
{
std::string str{"here is your string\n"};
return std::move(str); // <----- cast here?
}
int main ()
{
std::string my_string = std::move(get_string()); // <----- or cast here?
std::cout << my_string;
return 0;
}
So I have a simple example where I have a function that returns a copy of a string. I have read that its bad (and got the core-dumps to prove it!) to return any reference to a local temp variable so I have discounted trying that.
In the assignment in main() I don't want to copy-construct that string I want to move-construct/assign the string to avoid copying the string too much.
Q1: I return a "copy" of the temp var in get_string() - but I have cast the return value to rvalue-red. Is that pointless or is that doing anything useful?
Q2: Assuming Q1's answer is I don't need to do that. Then am I moving the fresh copy of the temp variable into my_string, or am I moving directly the temp variable str into my_string.
Q3: what is the minimum number of copies that you need in order to get a string return value stored into an "external" (in my case in main()) variable, and how do you do that (if I am not already achieving it)?
I return a "copy" of the temp var in get_string() - but I have cast the return value to rvalue-red. Is that pointless or is that doing anything useful?
You don't have to use std::move in that situation, as local variables returned by value are "implicitly moved" for you. There's a special rule in the Standard for this. In this case, your move is pessimizing as it can prevent RVO (clang warns on this).
Q2: Assuming Q1's answer is I don't need to do that. Then am I moving the fresh copy of the temp variable into my_string, or am I moving directly the temp variable str into my_string.
You don't need to std::move the result of calling get_string(). get_string() is a prvalue, which means that the move constructor of my_string will automatically be called (pre-C++17). In C++17 and above, mandatory copy elision will ensure that no moves/copies happen (with prvalues).
Q3: what is the minimum number of copies that you need in order to get a string return value stored into an "external" (in my case in main()) variable, and how do you do that (if I am not already achieving it)?
Depends on the Standard and on whether or not RVO takes place. If RVO takes place, you will have 0 copies and 0 moves. If you're targeting C++17 and initializing from a prvalue, you are guaranteed to have 0 copies and 0 moves. If neither take place, you'll probably have a single move - I don't see why any copy should occur here.
You do not need to use std::move on the return value which is a local variable. The compiler does that for you:
If expression is an lvalue expression that is the (possibly parenthesized) name of an automatic storage duration object declared in the body or as a parameter of the innermost enclosing function or lambda expression, then overload resolution to select the constructor to use for initialization of the returned value is performed twice: first as if expression were an rvalue expression (thus it may select the move constructor), and if no suitable conversion is available, or if the type of the first parameter of the selected constructor is not an rvalue reference to the object's type (possibly cv-qualified), overload resolution is performed a second time, with expression considered as an lvalue (so it may select the copy constructor taking a reference to non-const).
A particular property of c++'s lambda expressions is to capture the variables in the scope in which they are declared. For example I can use a declared and initialized variable c in a lambda function even if 'c' is not sent as an argument, but it's captured by '[ ]':
#include<iostream>
int main ()
{int c=5; [c](int d){std::cout<<c+d<<'\n';}(5);}
The expected output is thus 10. The problem arises when at least 2 variables, one captured and the other sent as an argument, have the same name:
#include<iostream>
int main ()
{int c=5; [c](int c){std::cout<<c<<'\n';}(3);}
I think that the 2011 standard for c++ says that the captured variable has the precedence on the arguments of the lambda expression in case of coincidence of names. In fact compiling the code using GCC 4.8.1 on Linux the output I get is the expected one, 5. If I compile the same code using apple's version of clang compiler (clang-503.0.40, the one which comes with Xcode 5.1.1 on Mac OS X 10.9.4) I get the other answer, 3.
I'm trying to figure why this happens; is it just an apple's compiler bug (if the standard for the language really says that the captured 'c' has the precedence) or something similar? Can this issue be fixed?
EDIT
My teacher sent an email to GCC help desk, and they answered that it's clearly a bug of GCC compiler and to report it to Bugzilla. So Clang's behavior is the correct one!
From my understanding of the c++11 standard's points below:
5.1.2 Lambda expressions
3 The type of the lambda-expression (which is also the type of the closure object) is a unique, unnamed non-union class type — called
the closure type — whose properties are described below.
...
5 The closure type for a lambda-expression has a public inline function call operator (13.5.4) whose parameters and return type are
described by the lambda-expression’s parameter-declaration-clause and
trailing-return-type respectively. This function call operator is
declared const (9.3.1) if and only if the lambda-expression’s
parameter-declaration-clause is not followed by mutable.
...
14 For each entity captured by copy, an unnamed non static data member is declared in the closure type
A lambda expression like this...
int c = 5;
[c](int c){ std::cout << c << '\n'; }
...is roughly equivalent to a class/struct like this:
struct lambda
{
int c; // captured c
void operator()(int c) const
{
std::cout << c << '\n';
}
};
So I would expect the parameter to hide the captured member.
EDIT:
In point 14 from the standard (quoted above) it would seem the data member created from the captured variable is * unnamed *. The mechanism by which is it referenced appears to be independent of the normal identifier lookups:
17 Every id-expression that is an odr-use (3.2) of an entity captured by copy is transformed into an access to the corresponding unnamed data member of the closure type.
It is unclear from my reading of the standard if this transformation should take precedence over parameter symbol lookup.
So perhaps this should be marked as UB (undefined behaviour)?
From the C++11 Standard, 5.1.2 "Lambda expressions" [expr.prim.lambda] #7:
The lambda-expression’s compound-statement yields the function-body (8.4) of the function call operator,
but for purposes of name lookup (3.4), determining the type and value of this (9.3.2) and transforming id-expressions
referring to non-static class members into class member access expressions using (*this) (9.3.1),
the compound-statement is considered in the context of the lambda-expression.
Also, from 3.3.3 "Block scope" [basic.scope.local] #2:
The potential scope of a function parameter name (including one appearing in a lambda-declarator) or of
a function-local predefined variable in a function definition (8.4) begins at its point of declaration.
Names in a capture list are not declarations and therefore do not affect name lookup. The capture list just allows you to use the local variables; it does not introduce their names into the lambda's scope. Example:
int i, j;
int main()
{
int i = 0;
[](){ i; }; // Error: Odr-uses non-static local variable without capturing it
[](){ j; }; // OK
}
So, since the parameters to a lambda are in an inner block scope, and since name lookup is done in the context of the lambda expression (not, say, the generated class), the parameter names indeed hide the variable names in the enclosing function.
I wonder how to make a variable within IDA Pro bound to some function so the next time I double click the variable it will send me to the function.
v1 = this
*v2 = Known-Function
At Some Different location:
char __stdcall ClassA__KnownFunction(ClassA *ClassA, void a2) {
commands.....
}
I know you can set type to int, struct, dword etc. But I am looking for some method to point the variable to already known offset/function in IDA Pro.
Function pointer is merely a variable that holds the address of a function; you cannot treat a variable like a constant. You have two options:
Add the name of the function as a comment (just for the sake of documentation).
Get rid of the variable assignment, hard-code the function address by editing the hex, and then perform the analysis again.
I've seen classes where constants are passed to methods, I guess its done to define some kind of setting in that function. I cant find it anywhere now to try to find out the logic, so I though I could ask here. How and why do you use this concept and where can I find more information about it?
The example below is written in PHP, but any language that handles constants would do I guess..
// Declaring class
class ExampleClass{
const EXAMPLE_CONST_1 = 0;
const EXAMPLE_CONST_2 = 1;
function example_method($constant(?)){
if($constant == ExampleClass::EXAMPLE_CONST_1)
// do this
else if($constant == ExampleClass::EXAMPLE_CONST_2)
// do that
}
}
// Using class
$inst = new ExampleClass();
$inst->example_method(ExampleClass::EXAMPLE_CONST_1);
To me its more clear to pass "ExampleClass::EXAMPLE_CONST_1" than to just pass "1", but it's that the only reason to pass constant?
Simply passing 1 doesn't say much. By having a constant you can have a description about the settings in the name.
example:
constant RAIN = 1;
method setWeather(RAIN);
Atleast that's how and why I use it.
It is always a good idea to avoid literals being passed around. By assigning a name, anyone reading your code has a chance to understand what that value means - a number has no meaning. It might also help you maintaining your code: If for some requirement the value has to be changed, you can easily do it in one place, instead of checking each and every value occurrence.