simple question for implementing a c++ operation in Fortran

[ya0037]

Hi,

Is there any one familiar with c++ who can help me implement this one in Fortran

static const unsigned int d1=0xDEADBEEF ^ (0x1fc4ce47*(d^(d>>13)));

Is the below one is the correct implementation:

Integer :: d1, d
Integer, Parameter :: cns1=Z'DEADBEEF'
Integer, Parameter :: cns2=Z'1fc4ce47'

d1=Ieor(cns1, cns2*Ieor(d, Ishft(d, -13)))

The above code gives me error the precision is not enough, but why?
If I change it to Integer(Kind = 16) it will work, but I do not understand in the c++ version they have used a simple 4 byte integer and that is enough, but in FORTRAN I have to use 16 bytes ????????????

Please some one give me a hint or st.

Regards,

 

[FJacq]

First of all, what is the value of "d" ? I don't see its declaration.

Normally all is OK if "d" is also declared INTEGER.


François Jacq

 

[ya0037]

yes d is unsigned int

But my program in FORTRAN gives error, so it is not OK:

Error: Arithmetic overflow converting INTEGER(16) to INTEGER(4)

Please tell me what should I do, if I want to convert it to higher precision I will come with a lot of problem, should I do that is there any solution?

Thank you very much.

 

[ya0037]

This is the whole error.
Please tell me what should I do?

Integer, Parameter :: cns1=Z'DEADBEEF'
                            1
Error: Arithmetic overflow converting INTEGER(16) to INTEGER(4) at (1). This check can be disabled with the option -fno-range-check

regards,

 

[FJacq]

I checked your program with ifort. It works correctly.
But effectively, gfortran issues an error message :

t47.f90:3.27:

Integer, Parameter :: cns1=Z'DEADBEEF'
                           1
Error: Arithmetic overflow converting INTEGER(16) to INTEGER(4) at (1). This check can be disabled with the option -fno-range-check
t47.f90:6.8:

d1=Ieor(cns1, cns2*Ieor(d, Ishft(d, -13)))
        1
Error: 'i' argument of 'ieor' intrinsic at (1) must be INTEGER

Of course, only the first message has to be examined (the second one is just a consequence of the first one). This message is due to the fact that Z'DEADBEEF', which is a positive binary integer, cannot be represented by a positive integer when an INTEGER(4) is used.

On the C side, you use an unsigned int. Such unsigned integer does not exist in Fortran.

You can follow the advice given by the compiler : add the option -fno-range-check. Your program compiles perfectly with this option.

Else, you have the possibility to modify slightly your program to make it standard conforming :

program test

integer,parameter :: long=selected_int_kind(10)
Integer :: d1, d=5
Integer(long), Parameter :: cns1=Z'DEADBEEF'
Integer(long), Parameter :: cns2=Z'1fc4ce47'

d1=Ieor(cns1, cns2*Ieor(d, Ishft(d, -13)))
write(*,*) d1
end program

As d1 is positive here, only the intermediate constants have to be declared "long".




François Jacq