I have this function, depicted below. It passes in two vectors with three values each, and should pass out one vector with three values as well. I call the function like this:
Fr = Flux(W(:,i),W(:,i+1))
What I have realized through messing around with the code, trying pure functions, and modules, and researching the error statement (that I will include at the bottom), is that fortran is reading my function Flux, and thinks that the input vectors are an attempt to call an entry from the array. That is my best guess as to what is going on. I asked around the lab and most people suggested using subroutines, but that seemed clunky, and I figured there should probably be a more elegant way, but I have not yet found it. I tried to define a result by saying:
DOUBLE PRECISION FUNCTION Flux(W1,W2) Result(FluxArray(3))
and then returning fluxarray but that does not work, as fortran cannot understand the syntax
The actual Function is this:
DOUBLE PRECISION FUNCTION Flux(W1,W2)
USE parameters
IMPLICIT NONE
DOUBLE PRECISION, DIMENSION(3), INTENT(IN)::W1, W2
DOUBLE PRECISION, DIMENSION(3), INTENT(OUT):: Flux
DOUBLE PRECISION, DIMENSION(3):: F1, F2
DOUBLE PRECISION::U1,U2,Rh1,Rh2,P1,P2,E1,E2,Rh,P,u,c,Lambda
INTEGER:: k
U1=W1(2)/W1(1)
U2=W2(2)/W2(1)
Rh1=W1(1)
Rh2=W2(1)
P1=(gamma_constant-1.d0)*(W1(3)-.5d0*Rh1*U1**2)
P2=(gamma_constant-1.d0)*(W2(3)-.5d0*Rh2*U2**2)
E1=W1(3)
E2=W2(3)
F1=[Rh1*U1,Rh1*U1**2+P1,(E1+P1)*U1]
F2=[Rh2*U2,Rh2*U2**2+P2,(E2+P2)*U2]
Rh=.5d0*(Rh1+Rh2)
P=.5d0*(P1+P2)
u=.5d0*(U1+U2)
c=sqrt(gamma_constant*P/Rh)
Lambda=max(u, u+c, u-c)
do k=1,3,1
Flux(k)=.5d0*(F1(k)+F2(k))-.5d0*eps*Lambda*(W2(k)-W1(k))
end do
RETURN
END FUNCTION Flux
Here is the error statement:
Quasi1DEuler.f90:191.51:
DOUBLE PRECISION, DIMENSION(3), INTENT(OUT):: Flux
1
Error: Symbol 'flux' at (1) already has basic type of REAL
Quasi1DEuler.f90:217.58:
Flux(k)=.5d0*(F1(k)+F2(k))-.5d0*eps*Lambda*(W2(k)-W1(k))
1
Error: Unexpected STATEMENT FUNCTION statement at (1)
Quasi1DEuler.f90:76.18:
Fr = Flux(W(:,i),W(:,i+1))
The last error occurs for both Fr and Fl. Thank you for your time and any help or consideration you can give!
EDIT/Follow-up::
Thanks for the help, I don't know a better way to present this so I'm going to edit the initial question.
I did as you suggested and It solved that issue, now it says:
Fr = Flux(W(:,i),W(:,i+1))
1
Error: The reference to function 'flux' at (1) either needs an explicit INTERFACE or the rank is incorrect
I saw a similar issue on SO at this link:
Computing the cross product of two vectors in Fortran 90
where they suggested that he put all his functions into modules. is there a better/simpler way for me to fix this error?
With RESULT(FluxArray), fluxArray is the name of the result variable. As such, your attempt to declare the characteristics in the result clause are mis-placed.
Instead, the result variable should be specified within the function body:
function Flux(W1,W2) result(fluxArray)
double precision, dimension(3), intent(in)::W1, W2
double precision, dimension(3) :: fluxArray ! Note, no intent for result.
end function Flux
Yes, one can declare the type of the result variable in the function statement, but the array-ness cannot be declared there. I wouldn't recommend having a distinct dimension statement in the function body for the result variable.
Note that, when referencing a function returning an array it is required that there be an explicit interface available to the caller. One way is to place the function in a module which is used. See elsewhere on SO, or language tutorials, for more details.
Coming to the errors from your question without the result.
DOUBLE PRECISION FUNCTION Flux(W1,W2)
DOUBLE PRECISION, DIMENSION(3), INTENT(OUT):: Flux
Here the type of Flux has been declared twice. Also, it isn't a dummy argument to the function, so as above it need not have the intent attribute.
One could write
FUNCTION Flux(W1,W2)
DOUBLE PRECISION, DIMENSION(3) :: Flux ! Deleting intent
or (shudder)
DOUBLE PRECISION FUNCTION Flux(W1,W2)
DIMENSION :: Flux(3)
The complaint about a statement function is unimportant here, following on from the bad declaration.
Thank you in advance for your time and help.
Exercise:
Write the code for sumDigitsInNumber(int number). The method takes a three-digit whole number. You need to calculate the sum of the digits of this number, and then return the result.
Consider this example:
The sumDigitsInNumber method is called with argument 546.
Example output:
15
CODE:
public class Solution {
public static void main(String[] args) {
System.out.println(sumDigitsInNumber(546));
}
public static int sumDigitsInNumber(int number) {
return number ==0? 0:number%10+sumDigitsInNumber(number/10);
}
}
This is a solution and the task has been passed. The problem is the solution had been implemented by someone (not by me) therefore I can't understand How this function does its job.
I tried to test the function parts separately, just to see what would happen, and here is the result:
number%10 = 546%10;
546/10 = 54;
output:
6+sumDigitsInNumber(546/10) - which is totally wrong.
I don't understand HOW sumDigitsInNumber is treated by the ternary operator in there and how this short line of code:
return number ==0? 0:number%10+sumDigitsInNumber(number/10);
makes such a complicated calculation?
Can anyone explain it to me in a way it would have explained to a Java-child?
TYVM in advance.
So, using the example number of 546, let's step through the code.
In the first run, it does indeed return 6+sumDigitsInNumber(546/10), that is all correct.
Because sumDigitsInNumber's parameter (number) is int, the decimal portion of the division is truncated, resulting in essentially a floor operation (forced round down). And we recursively call sumDigitsInNumber's, so we just keep "looping" that section of code. So for the second run, it is equivalent to sumDigitsInNumber(54), plus the additional 6 from the first run (6+sumDigitsInNumber(54)).
The second call returns 4+sumDigitsInNumber(54/10) by following the same logic as the first call. This is equivalent to 4+sumDigitsInNumber(5).
Then we run the whole process again, which returns 5+sumDigitsInNumber(5/10), equivalent to 5+sumDigitsInNumber(0).
The final call, sumDigitsInNumber(0), will return 0 because of the ternary operator in the return statement.
To expand this all out:
sumDigitsInNumber(546)
= 6+sumDigitsInNumber(546/10) = 6+sumDigitsInNumber(54)
= 6+(4+sumDigitsInNumber(54/10)) = 6+(4+sumDigitsInNumber(5))
= 6+(4+(5+sumDigitsInNumber(5/10))) = 6+(4+(5+sumDigitsInNumber(0)))
= 6+(4+(5+0))
= 6+(4+(5))
= 6+(9)
= 15
I tried to calculate an exponential calculation, I tried both ways, writing it with the normal ** operator and the pow() function.
If I directly use numbers, everything works completely fine. But as soon as I use variables which get their value from input() functions the result is rounded, although I use the float argument.
I am quite new to coding, so please don't go to hard on me.
Code below:
pow(1.05,5), everything fine, result is 1.2762815625 and so on
float(pow(int(a),int(b)) the result is just 1.0, although it should be the same as above.
it is because you are doing the power operation on int i.e first you are converting the 1.05 to 1 by doing int(1.05) and then calculating the power(). you need to apply pow() on float()
print(float(pow(float(1.05),float(5))))
or either
print(pow(float(1.05),float(5)))
The reason the float function returns 1.0 is because of how it deals with integers.
For example:
integer = 5.5236734
print(float(integer))
The above code's output will be:
5.5236734
Now lets say you make some changes:
integer = 5.5236734
print(int(integer))
First, we made integer a decimal number, and then we said to print the int() form of the decimal.
This will be the result:
5
So, to fix your code, you just need to do this:
a = 1.05
b = 5
print(float(pow(a, b)))
Which will output:
1.2762815625000004
Hope this helps!
I am using DataListView from Bright Idea Software's ObjectListView to show real time data. I need to show double values correct upto 4 decimal point. How can I implement the same?
I suppose you currently use AspectName property to get/set value. Use the AspectGetter instead and format the return value as required.
Assuming you have an model object of type "Item" with a property "DoubleValue" of type double:
olvColumn1.AspectGetter += delegate(object rowObject) {
Item item = rowObject as Item;
return Math.Round(item.DoubleValue, 4);
};
You could also convert the DoubleValue using ToString(), but that would only be advisable if you do not need to edit the property from the OLV. Because the OLV "sees" the type you return from the AspectGetter (which would then be string instead of double) and not use a NumericUpDown control if you would try to edit the value.
I try to get an integer (right after "PLAYING:STATION\nID:"out of the String shown in my screenshot by using the following code:
int zahl = Integer.parseInt(sentence.substring(sentence.indexOf("PLAYING:STATION\n
ID:")+1).trim());
But all I get is a NumberFormatException. How can I tell the trim()-method that it has to stop right after the number?
Screenshot of the complete String
Since you must select a specific ID: entry among several ones, I think the best way to proceed is using a regular expression. Using your output sample, I wrote the following code:
String text = "PLAYING_MODE\nID:\nPLAYING:STATION\nID:2\nNAME:wazee.org";
Pattern p = Pattern.compile("PLAYING:STATION\nID:(\\d+)");
Matcher m = p.matcher(text);
if (m.find()) {
int number = Integer.parseInt(m.group(1));
System.out.println(number);
}
which correctly parses the ID number that immediately follows PLAYING_STATION.
You could as well repeatedly work with the overloaded String.indexOf() method (find PLAYING:STATION, then the following ID:, then the following \n). I think the code might be harder to read, but it would still do the job.
I hope this will be helpful...
Cheers,
Jeff