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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.
How come that in the following snippet
int a = 7;
int b = 3;
double c = 0;
c = a / b;
c ends up having the value 2, rather than 2.3333, as one would expect. If a and b are doubles, the answer does turn to 2.333. But surely because c already is a double it should have worked with integers?
So how come int/int=double doesn't work?
This is because you are using the integer division version of operator/, which takes 2 ints and returns an int. In order to use the double version, which returns a double, at least one of the ints must be explicitly casted to a double.
c = a/(double)b;
Here it is:
a) Dividing two ints performs integer division always. So the result of a/b in your case can only be an int.
If you want to keep a and b as ints, yet divide them fully, you must cast at least one of them to double: (double)a/b or a/(double)b or (double)a/(double)b.
b) c is a double, so it can accept an int value on assignement: the int is automatically converted to double and assigned to c.
c) Remember that on assignement, the expression to the right of = is computed first (according to rule (a) above, and without regard of the variable to the left of =) and then assigned to the variable to the left of = (according to (b) above). I believe this completes the picture.
With very few exceptions (I can only think of one), C++ determines the
entire meaning of an expression (or sub-expression) from the expression
itself. What you do with the results of the expression doesn't matter.
In your case, in the expression a / b, there's not a double in
sight; everything is int. So the compiler uses integer division.
Only once it has the result does it consider what to do with it, and
convert it to double.
When you divide two integers, the result will be an integer, irrespective of the fact that you store it in a double.
c is a double variable, but the value being assigned to it is an int value because it results from the division of two ints, which gives you "integer division" (dropping the remainder). So what happens in the line c=a/b is
a/b is evaluated, creating a temporary of type int
the value of the temporary is assigned to c after conversion to type double.
The value of a/b is determined without reference to its context (assignment to double).
In C++ language the result of the subexpresison is never affected by the surrounding context (with some rare exceptions). This is one of the principles that the language carefully follows. The expression c = a / b contains of an independent subexpression a / b, which is interpreted independently from anything outside that subexpression. The language does not care that you later will assign the result to a double. a / b is an integer division. Anything else does not matter. You will see this principle followed in many corners of the language specification. That's juts how C++ (and C) works.
One example of an exception I mentioned above is the function pointer assignment/initialization in situations with function overloading
void foo(int);
void foo(double);
void (*p)(double) = &foo; // automatically selects `foo(fouble)`
This is one context where the left-hand side of an assignment/initialization affects the behavior of the right-hand side. (Also, reference-to-array initialization prevents array type decay, which is another example of similar behavior.) In all other cases the right-hand side completely ignores the left-hand side.
The / operator can be used for integer division or floating point division. You're giving it two integer operands, so it's doing integer division and then the result is being stored in a double.
This is technically a language-dependent, but almost all languages treat this subject the same. When there is a type mismatch between two data types in an expression, most languages will try to cast the data on one side of the = to match the data on the other side according to a set of predefined rules.
When dividing two numbers of the same type (integers, doubles, etc.) the result will always be of the same type (so 'int/int' will always result in int).
In this case you have
double var = integer result
which casts the integer result to a double after the calculation in which case the fractional data is already lost. (most languages will do this casting to prevent type inaccuracies without raising an exception or error).
If you'd like to keep the result as a double you're going to want to create a situation where you have
double var = double result
The easiest way to do that is to force the expression on the right side of an equation to cast to double:
c = a/(double)b
Division between an integer and a double will result in casting the integer to the double (note that when doing maths, the compiler will often "upcast" to the most specific data type this is to prevent data loss).
After the upcast, a will wind up as a double and now you have division between two doubles. This will create the desired division and assignment.
AGAIN, please note that this is language specific (and can even be compiler specific), however almost all languages (certainly all the ones I can think of off the top of my head) treat this example identically.
For the same reasons above, you'll have to convert one of 'a' or 'b' to a double type. Another way of doing it is to use:
double c = (a+0.0)/b;
The numerator is (implicitly) converted to a double because we have added a double to it, namely 0.0.
The important thing is one of the elements of calculation be a float-double type. Then to get a double result you need to cast this element like shown below:
c = static_cast<double>(a) / b;
or
c = a / static_cast(b);
Or you can create it directly::
c = 7.0 / 3;
Note that one of elements of calculation must have the '.0' to indicate a division of a float-double type by an integer. Otherwise, despite the c variable be a double, the result will be zero too (an integer).
I'm trying to crate a function in Fortran (95) that that will have as input a string (test) and a character (class). The function will compare each character of test with the character class and return a logical that is .true. if they are of the same class1 and .false. otherwise.
The function (and the program to run it) is defined below:
!====== WRAPPER MODULE ======!
module that_has_function
implicit none
public
contains
!====== THE ACTUAL FUNCTION ======!
function isa(test ,class )
implicit none
logical, allocatable, dimension(:) :: isa
character*(*) :: test
character :: class
integer :: lt
character(len=:), allocatable :: both
integer, allocatable, dimension(:) :: intcls
integer :: i
lt = len_trim(test)
allocate(isa(lt))
allocate(intcls(lt+1))
allocate(character(len=lt+1) :: both)
isa = .false.
both = class//trim(test)
do i = 1,lt+1
select case (both(i:i))
case ('A':'Z'); intcls(i) = 1! uppercase alphabetic
case ('a':'a'); intcls(i) = 2! lowercase alphabetic
case ('0':'9'); intcls(i) = 3! numeral
case default; intcls(i) = 99! checks if they are equal
end select
end do
isa = intcls(1).eq.intcls(2:)
return
end function isa
end module that_has_function
!====== CALLER PROGRAM ======!
program that_uses_module
use that_has_function
implicit none
integer :: i
i = 65
! Reducing the result of "isa" to a scalar with "all" works:
! V-V
do while (all(isa(achar(i),'A')))
print*, achar(i)
i = i + 1
end do
! Without the reduction it doesn''t:
!do while (isa(achar(i),'A'))
! print*, achar(i)
! i = i + 1
!end do
end program that_uses_module
I would like to use this function in do while loops, for example, as it is showed in the code above.
The problem is that, for example, when I use two scalars (rank 0) as input the function still returns the result as an array (rank 1), so to make it work as the condition of a do while loop I have to reduce the result to a scalar with all, for example.
My question is: can I make the function conditionally return a scalar? If not, then is it possible to make the function work with vector and scalar inputs and return, respectively, vector and scalar outputs?
1. What I call class here is, for example, uppercase or lowercase letters, or numbers, etc. ↩
You can not make the function conditionally return a scalar or a vector.
But you guessed right, there is a solution. You will use a generic function.
You write 2 functions, one that takes scalar and return scalar isas, the 2nd one takes vector and return vector isav.
From outside of the module you will be able to call them with the same name: isa. You only need to write its interface at the beginning of the module:
module that_has_function
implicit none
public
interface isa
module procedure isas, isav
end interface isa
contains
...
When isa is called, the compiler will know which one to use thanks to the type of the arguments.
The rank of a function result cannot be conditional on the flow of execution. This includes selection by evaluating an expression.
If reduction of a scalar result is too much, then you'll probably be horrified to see what can be done instead. I think, for instance, of derived types and defined operations.
However, I'd consider it bad design in general for the function reference to be unclear in its rank. My answer, then, is: no you can't, but that's fine because you don't really want to.
Regarding the example of minval, a few things.1 As noted in the comment, minval may take a dim argument. So
integer :: X(5,4) = ...
print *, MINVAL(X) ! Result a scalar
print *, MINVAL(X,dim=1) ! Result a rank-1 array
is in keeping with the desire of the question.
However, the rank of the function result is still "known" at the time of referencing the function. Simply having a dim argument means that the result is an array of rank one less than the input array rather than a scalar. The rank of the result doesn't depend on the value of the dim argument.
As noted in the other answer, you can have similar functionality with a generic interface. Again, the resolved specific function (whichever is chosen) will have a result of known rank at the time of reference.
1 The comment was actually about minloc but minval seems more fitting to the topic.
In RcppArmadillo, I need to know how I can convert arma::mat to c-style array double * for use in other functions.
When I run the following functions, the computer crashes:
R part:
nn3 <- function(x){
results=.Call("KNNCV", PACKAGE = "KODAMA", x)
results
}
C++ part:
double KNNCV(arma::mat x) {
double *cvpred = x.memptr();
return cvpred[1];
}
and at the end, I try:
nn3(as.matrix(iris[,-5]))
Can you help me to find the errors, please?
First, there is no such such thing as vector stored in a double*. You can cast to a C-style pointer to double; but without length information that does not buy you much.
By convention, most similar C++ classes give you a .begin() iterator to the beginning of the memory block (which Armadillo happens to guarantee to be contiguous, just like std::vector) so you can try that.
Other than that the (very fine indeed) Armadillo documentation tells you about memptr() which is probably what you want here. Straight copy from the example there:
mat A = randu<mat>(5,5);
const mat B = randu<mat>(5,5);
double* A_mem = A.memptr();
const double* B_mem = B.memptr();
I am trying to produce the argument string for an anonymous function based on the number of input arguments without using for loops. For example, if N=3, then I want a string that reads
#(ax(1),ax(2),ax(3),ay(1),ay(2),ay(3))
I tried using repmat('ax',1,N) but I cannot figure out how to interleave the (i) index.
Any ideas?
Aside: Great answers so far, the above problem has been solved. To provide some intuition for those who are wondering why I want to do this: I need to construct a very large matrix anonymous function (a Jacobian) on the order of 3000x3000. I initially used the Matlab operations jacobian and matlabFunction to construct the anonymous function; however, this was quite slow. Instead, since the closed form of the derivative was quite simple, I decided to form the anonymous function directly. This was done by forming the symbolic Jacobian matrix, J, then appending it to the above #() string by using char(J{:})' and using eval to form the final anonymous function. This may not be the most elegant solution but I find it runs much faster than the jacobian/matlabFunction combination, especially for large N (additionally the structure of the new approach allows for the evaluation to be done in parallel).
EDIT: Just for completeness, the correct form of the argument string for the anonymous function should read
#(ax1,ax2,ax3,ay1,ay2,ay3)
to avoid a syntax error associated with indexing.
I suggest the following:
N = 3;
argumentString = [repmat('ax(%i),',1,N),repmat('ay(%i),',1,N)];
functionString = sprintf(['#(',argumentString(1:end-1),')'], 1:N, 1:N)
First, you create input masks for sprintf (e.g. 'ax(%i)'), which you then fill in with the appropriate numbers to create the function string.
Note: the syntax #(ax(1),...) will not actually work. More likely, you want to use either #()someFunction(ax(1),...), or you are trying to pass multiple input arguments to an existing function, in which case storing the inputs in a cell array and calling the function as fun(axCell{:}) would work.
A solution would be to use arrayfun:
sx = strjoin(arrayfun(#(x) ['ax(' num2str(x) ')'], 1:3, 'UniformOutput', false), ',');
sy = strjoin(arrayfun(#(x) ['ay(' num2str(x) ')'], 1:3, 'UniformOutput', false), ',');
s = ['#(' sx ',' sy ')'];
contains
'#(ax(1),ax(2),ax(3),ay(1),ay(2),ay(3))'
Best,
Try this:
N = 3;
sx = strcat('ax(', arrayfun(#num2str, 1:N, 'uniformoutput', 0), '),');
sy = strcat('ay(', arrayfun(#num2str, 1:N, 'uniformoutput', 0), '),');
str = [sx{:} sy{:}];
str = ['#(' str(1:end-1) ')']