Does anyone know how to make Resharper add blank lines after brackets when performing code cleanup?
I want to make the following code
if (x == y)
{
int a = x;
int b = y;
int z = a + b;
}
Look like this
if (x == y)
{
int a = x;
int b = y;
int z = a + b;
}
I have had a look through the Resharper options but can't see anything to add the extra blank lines around your block of code, although I am not sure that format makes it any more readable than without extra spaces. I would suggest asking this question on the Resharper News Group: news://news.jetbrains.com/jetbrains.resharper.community
Related
Let's assume we have this code:
int x;
double yyyy;
float zzzzzzzz;
x = 1;
yyyy = 24.5d;
zzzzzzzz = 1.3f;
after I enabled:
Settings -> Code Style-> Java -> Wrapping and Braces -> Align Variable in columns
and when I press Ctrl+Alt+L it It is reformatted to the following:
int x;
double yyyy;
float zzzzzzzz;
x = 1;
yyyy = 24.5d;
zzzzzzzz = 1.3f;
Is there any way to make the result like this:
int x;
double yyyy;
float zzzzzzzz;
x = 1;
yyyy = 24.5d;
zzzzzzzz = 1.3f;
with equal indents when assignment multi variables like last three lines.
No way to align the code like that. After five years, there is no way still.
I've searched for an hour for the methods doing numerical integration. I'm new to Rcpp and rewriting my old programs now. What I have done in R was:
x=smpl.x(n,theta.true)
joint=function(theta){# the joint dist for
#all random variable
d=c()
for(i in 1:n){
d[i]=den(x[i],theta)
}
return(prod(d)*dbeta(theta,a,b)) }
joint.vec=Vectorize(joint)##vectorize the function, as required when
##using integrate()
margin=integrate(joint.vec,0,1)$value # the
##normalizeing constant at the donominator
area=integrate(joint.vec,0,theta.true)$value # the values at the
## numeritor
The integrate() function in R will be slow, and since I am doing the integration for a posterior distribution of a sample of size n, the value of the integration will be huge with large error.
I am trying to rewrite my code with the help of Rcpp, but I　don't know how to deal with the integrate. Should I include a c++ h file? Or any suggestions?
You can code your function in C and call it, for instance, via the sourceCpp function and then integrate it in R. In alternative, you can call the integrate function of R within your C code by using the Function macro of Rcpp. See Dirk's book (Seamless R and C++ Integration with Rcpp) on page 56 for an example of how to call R functions from C. Another alternative (which I believe is the best for most cases) is to integrate your function written in C , directly in C, using the RcppGSL package.
As about the huge normalizing constant, sometimes it is better to scale the function at the mode before integrating it (you can find modes with, e.g., nlminb, optim, etc.). Then, you integrate the rescaled function and to recover the original nroming constant multiply the resulting normalizing constant by the rescaling factor. Hope this may help!
after reading your #utobi advice, I felt programming by my own maybe easier. I simply use Simpson formula to approximate the integral:
// [[Rcpp::export]]
double den_cpp (double x, double theta){
return(2*x/theta*(x<=theta)+2*(1-x)/(1-theta)*(theta<x));
}
// [[Rcpp::export]]
double joint_cpp ( double theta,int n,NumericVector x, double a, double b){
double val = 1.0;
NumericVector d(n);
for (int i = 0; i < n; i++){
double tmp = den_cpp(x[i],theta);
val = val*tmp;
}
val=val*R::dbeta(theta,a,b,0);
return(val);
}
// [[Rcpp::export]]
List Cov_rate_raw ( double theta_true, int n, double a, double b,NumericVector x){
//This function is used to test, not used in the fanal one
int steps = 1000;
double s = 0;
double start = 1.0e-4;
std::cout<<start<<" ";
double end = 1-start;
std::cout<<end<<" ";
double h = (end-start)/steps;
std::cout<<"1st h ="<<h<<" ";
double area = 0;
double margin = 0;
for (int i = 0; i < steps ; i++){
double at_x = start+h*i;
double f_val = (joint_cpp(at_x,n,x,a,b)+4*joint_cpp(at_x+h/2,n,x,a,b)+joint_cpp(at_x+h,n,x,a,b))/6;
s = s + f_val;
}
margin = h*s;
s=0;
h=(theta_true-start)/steps;
std::cout<<"2nd h ="<<h<<" ";
for (int i = 0; i < steps ; i++){
double at_x = start+h*i;
double f_val = (joint_cpp(at_x,n,x,a,b)+4*joint_cpp(at_x+h/2,n,x,a,b)+joint_cpp(at_x+h,n,x,a,b))/6;
s = s + f_val;
}
area = h * s;
double r = area/margin;
int cover = (r>=0.025)&&(r<=0.975);
List ret;
ret["s"] = s;
ret["margin"] = margin;
ret["area"] = area;
ret["ratio"] = r;
ret["if_cover"] = cover;
return(ret);
}
I'm not that good at c++, so the two for loops like kind of silly.
It generally works, but there are still several potential problems:
I don't really know how to choose the steps, or how many sub intervals do I need to approximate the integrals. I've taken numerical analysis when I was an undergraduate, I think maybe I need to check my book about the expression of the error term, to decide the step length.
I compared my results with those from R. the integrate() function in R can take care of the integral over the interval [0,1]. That helps me because my function is undefined at 0 or 1, which takes infinite value. In my C++ code, I can only make my interval from [1e-4, 1-1e-4]. I tried different values like 1e-7, 1e-10, however, 1e-4 was the one most close to R's results....What should I do with it?
Question: What is y after the following switch statement is executed? Rewrite the code using an if statement.
y = 3; x = 3;
switch (x + 3)
{
case 6: y = 1;
default: y += 1;
}
I'm at the beginning of the C++ hike. I don't know what do with this. It's not working in C++ Visual Studio 2013. I've put it in as is and nothing happens.
I use:
y = 3; x = 3;
switch (x + 3)
{
case 6: y = 1;
default: y += 1;
}
return 0;
}
And nothing happens. I have both the answers but I have no clue how to get them...
y is 2
if (x + 3 == 6)
y = 1;
y += 1;
I'm strictly supposed to be using
#include <iostream>
using namespace std;
int main()
{
return 0;
}
switch basics in C/C++:
switch expression is evaluated once, at the beginning. Then the 1st matching case is executed
after the case clause, the execution falls through to the next clause unless there's a break. Since this typically isn't what was indended, fall-throughs are normally documented with a comment or (if multiple cases refer to the same code) by stacking them together (case 2: case 3: <code>).
I try to understand lexical-scoping. In lexical-scoping, I have this code, C like syntax:
main{
f1(){
int y = 8;
}
int y = 1; //*
f1();
}
After the execution of f1() line, will the value of y variable in main (I put * next of it) remain 1 or change to 8?
It will remain 1. You have two completely distinct variables. Changes to one do not affect the other.
Does anyone know of a truly declarative language? The behavior I'm looking for is kind of what Excel does, where I can define variables and formulas, and have the formula's result change when the input changes (without having set the answer again myself)
The behavior I'm looking for is best shown with this pseudo code:
X = 10 // define and assign two variables
Y = 20;
Z = X + Y // declare a formula that uses these two variables
X = 50 // change one of the input variables
?Z // asking for Z should now give 70 (50 + 20)
I've tried this in a lot of languages like F#, python, matlab etc, but every time I tried this they come up with 30 instead of 70. Which is correct from an imperative point of view, but I'm looking for a more declarative behavior if you know what I mean.
And this is just a very simple calculation. When things get more difficult it should handle stuff like recursion and memoization automagically.
The code below would obviously work in C# but it's just so much code for the job, I'm looking for something a bit more to the point without all that 'technical noise'
class BlaBla{
public int X {get;set;} // this used to be even worse before 3.0
public int Y {get;set;}
public int Z {get{return X + Y;}}
}
static void main(){
BlaBla bla = new BlaBla();
bla.X = 10;
bla.Y = 20;
// can't define anything here
bla.X = 50; // bit pointless here but I'll do it anyway.
Console.Writeline(bla.Z);// 70, hurray!
}
This just seems like so much code, curly braces and semicolons that add nothing.
Is there a language/ application (apart from Excel) that does this? Maybe I'm no doing it right in the mentioned languages, or I've completely missed an app that does just this.
I prototyped a language/ application that does this (along with some other stuff) and am thinking of productizing it. I just can't believe it's not there yet. Don't want to waste my time.
Any Constraint Programming system will do that for you.
Examples of CP systems that have an associated language are ECLiPSe, SICSTUS Prolog / CP package, Comet, MiniZinc, ...
It looks like you just want to make Z store a function instead of a value. In C#:
var X = 10; // define and assign two variables
var Y = 20;
Func<int> Z = () => X + Y; // declare a formula that uses these two variables
Console.WriteLine(Z());
X = 50; // change one of the input variables
Console.WriteLine(Z());
So the equivalent of your ?-prefix syntax is a ()-suffix, but otherwise it's identical. A lambda is a "formula" in your terminology.
Behind the scenes, the C# compiler builds almost exactly what you presented in your C# conceptual example: it makes X into a field in a compiler-generated class, and allocates an instance of that class when the code block is entered. So congratulations, you have re-discovered lambdas! :)
In Mathematica, you can do this:
x = 10; (* # assign 30 to the variable x *)
y = 20; (* # assign 20 to the variable y *)
z := x + y; (* # assign the expression x+y to the variable z *)
Print[z];
(* # prints 30 *)
x = 50;
Print[z];
(* # prints 70 *)
The operator := (SetDelayed) is different from = (Set). The former binds an unevaluated expression to a variable, the latter binds an evaluated expression.
Wanting to have two definitions of X is inherently imperative. In a truly declarative language you have a single definition of a variable in a single scope. The behavior you want from Excel corresponds to editing the program.
Have you seen Resolver One? It's like Excel with a real programming language behind it.
Here is Daniel's example in Python, since I noticed you said you tried it in Python.
x = 10
y = 10
z = lambda: x + y
# Output: 20
print z()
x = 20
# Output: 30
print z()
Two things you can look at are the cells lisp library, and the Modelica dynamic modelling language, both of which have relation/equation capabilities.
There is a Lisp library with this sort of behaviour:
http://common-lisp.net/project/cells/
JavaFX will do that for you if you use bind instead of = for Z
react is an OCaml frp library. Contrary to naive emulations with closures it will recalculate values only when needed
Objective Caml version 3.11.2
# #use "topfind";;
# #require "react";;
# open React;;
# let (x,setx) = S.create 10;;
val x : int React.signal = <abstr>
val setx : int -> unit = <fun>
# let (y,sety) = S.create 20;;
val y : int React.signal = <abstr>
val sety : int -> unit = <fun>
# let z = S.Int.(+) x y;;
val z : int React.signal = <abstr>
# S.value z;;
- : int = 30
# setx 50;;
- : unit = ()
# S.value z;;
- : int = 70
You can do this in Tcl, somewhat. In tcl you can set a trace on a variable such that whenever it is accessed a procedure can be invoked. That procedure can recalculate the value on the fly.
Following is a working example that does more or less what you ask:
proc main {} {
set x 10
set y 20
define z {$x + $y}
puts "z (x=$x): $z"
set x 50
puts "z (x=$x): $z"
}
proc define {name formula} {
global cache
set cache($name) $formula
uplevel trace add variable $name read compute
}
proc compute {name _ op} {
global cache
upvar $name var
if {[info exists cache($name)]} {
set expr $cache($name)
} else {
set expr $var
}
set var [uplevel expr $expr]
}
main
Groovy and the magic of closures.
def (x, y) = [ 10, 20 ]
def z = { x + y }
assert 30 == z()
x = 50
assert 70 == z()
def f = { n -> n + 1 } // define another closure
def g = { x + f(x) } // ref that closure in another
assert 101 == g() // x=50, x + (x + 1)
f = { n -> n + 5 } // redefine f()
assert 105 == g() // x=50, x + (x + 5)
It's possible to add automagic memoization to functions too but it's a lot more complex than just one or two lines. http://blog.dinkla.net/?p=10
In F#, a little verbosily:
let x = ref 10
let y = ref 20
let z () = !x + !y
z();;
y <- 40
z();;
You can mimic it in Ruby:
x = 10
y = 20
z = lambda { x + y }
z.call # => 30
z = 50
z.call # => 70
Not quite the same as what you want, but pretty close.
not sure how well metapost (1) would work for your application, but it is declarative.
Lua 5.1.4 Copyright (C) 1994-2008 Lua.org, PUC-Rio
x = 10
y = 20
z = function() return x + y; end
x = 50
= z()
70
It's not what you're looking for, but Hardware Description Languages are, by definition, "declarative".
This F# code should do the trick. You can use lazy evaluation (System.Lazy object) to ensure your expression will be evaluated when actually needed, not sooner.
let mutable x = 10;
let y = 20;
let z = lazy (x + y);
x <- 30;
printf "%d" z.Value