why the result of CVF and CPP not equal

[Intel_C_Intel]

CVF and PGF90:
real*8 rs
rs=dcos(dsin(1100000.0))
! rs=0.927451417863706
CPP(g++):
double i;
i=1100000.0;
i=cos(sin(i));

// i=0.999999
what's the matter?
Thank!

[Intel_C_Intel]

FORTRAN:
real*8 x(10),y(10),i
integer j
x=(/(1.1*j,j=1,10)/)
do i=1.0,10000.0,0.1
y=dcos(dsin(x*i))
enddo
print*,y
print*,"---------------------------------------"
print*,"---------------------------------------"
x=(/(1.1d0*j,j=1,10)/)
do i=1.0d0,10000.0d0,0.1d0
y=dcos(dsin(x*i))
enddo
print*,y
end

Result:
0.603824893601652 0.756962151192525 0.930925120912308
0.540388345095186 0.920971720348514 0.769015433909829
0.595637597164346 0.999591124599520 0.611723873042041
0.742800902664028
---------------------------------------
---------------------------------------
0.574184213326437 0.856445471589781 0.794619918650350
0.612629194927815 0.990632673212321 0.549625052870082
0.912164220137757 0.731635495768517 0.662642071205611
0.963401666343262

CPP:
#include
#include
void main()
{
double x[11],y[11],i;
int j;
for(j=1;j<=10;j++) x=1.1*j;
for(i=double(1.0);i<=double(10000.0);i+=double(0.1))
{
for(j=1;j<=10;j++)
{
y=cos(sin(x*i));
}
}
for(j=1;j<=10;j++) printf("%f ",y);
getchar();
}

Result:
0.603949 0.756593 0.931736 0.540409 0.920679 0.771226 0.594482 0.999494 0.614068 0.742055

[Intel_C_Intel]

I think that CVF is getting the answer correct. In that case the question is (to ask on a C++ forum?) is why G++ is getting it wrong. This may narrow things down a bit.

Joe

[Intel_C_Intel]

Thank you!
Can you tell me wnhy you think so,just for you think my code is right?
But I think the c++ code is right too.

I will ask on c++ forum as you said.
thanks!

[david_jones]

It would be illuminating to (either or both):
(i) count in each case how many times the i-loop is actually transited;
(ii) print out, to as many dp as needed, the actual last value of i that is used.

A guess is that there will be differences because of the use of a non-integer loop control variable with a step that is not exactly computer-representable. You could potentially find that changing compiler options regarding code generation/optimisation would lead to different answers even keeping the code/language the same.

[Intel_C_Intel]

thank
i change my code:
f90:
do i=1.0d0,100.0d0,0.10000000d0
c++:
for(i=double(1.0);i<=double(100.0);i+=double(0.10000000))
the result of i on pgf90 or pgCC:

f90:
4.200000000000002
4.300000000000002
4.400000000000001
4.500000000000001
4.600000000000001
4.700000000000000
4.800000000000000
4.899999999999999
4.999999999999999
5.099999999999999
5.199999999999998
5.299999999999998

c++:
4.80000000000000
4.90000000000000
5.00000000000000
5.10000000000000
5.20000000000000
5.30000000000000
5.40000000000000
5.50000000000000
5.60000000000000
5.70000000000000
5.80000000000000
5.90000000000000
6.00000000000000
6.10000000000000
6.19999999999999
6.29999999999999
6.39999999999999
6.49999999999999
6.59999999999999
6.69999999999999
6.79999999999999

maybe this show the c++ better than f90 ?
:(

[david_jones]

You have given a different number of decimal places in the two cases ... the results agree to the smaller number of places. For your test the most important value to quote will be the one at the end of the loop, where any inaccuracies will have had the chance to build up and you may or may not be on the right side of the upper bound due to numerical inaccuracy. It is the last value in the loop that is used in your test example code and the value of i may differ by 0.1

What are pgf90 etc.? As I recall, CVF precomputes how many times it will go round the loop by dividing the range by the step-size. Possibly C++ implements the loop by increasing i incrementally and testing each time whether it is within the limit.

[Intel_C_Intel]

Thank you much!
I think you are right,thank!

i don't know what you mean about 'what are pgf90',pgf90 is a very good f90 and cpp compiler on Linux.As i get the same result with pgi90,I think it just because of the fortran itself not of CVF.

http://www.pgroup.com/