here is my program:
real a(2,240)
integer i,j
if (TIME(2).EQ.0) then
!读入温度数据
open(12,file='temperature.txt')
do i=1,240
read(12,*) a(:,i)
print*, a(:,i)
end do
close(12)
end if
! 对应时间步长
do j = 1,240
if (TIME(2).EQ.a(1,j)) then
SINK = a(2,j)
exit
else
SINK = 0
end if
end do
but in the job.log i find the result of print is all zero. why!?
here is my txt:
0 28.7100000000000
0.500000000000000 28.3600000000000
1 28.0300000000000
1.50000000000000 27.6100000000000
2 27.4900000000000
2.50000000000000 27.0800000000000
3 26.7400000000000
3.50000000000000 26.7200000000000
....
119.500000000000 24.3700000000000
120 24.3300000000000
I can see several problems with this code plus I have a *real* FORTRAN compiler (not just some silly LINUX thing). IF(TIME(2).EQ.0) is a problem if TIME() isn't an integer (never compare a floating-point to an integer and never expect a floating-point to exactly equal any particular number, always use .LE.0.0 or something like that). Read a(:,i) is a problem too. If you mean a(1,i) then put that. If you mean read(12,*)(a(j,i),j=1,2) then put that.
thanks for your kindly help, and i have learnt a lot from your reply, i should improve my code more seriously.
I copied the data plus filled in the missing range (4 to 119) with something reasonable. The data are in temperature.txt Then I pasted the code into PROG.FOR, compiled it with Digital FORTRAN, and ran it. The compiler messages and program output are in output.txt. I included the program (PROG.FOR) and the compiled Windows 32-bit executable (PROG.exe) in the attached archive. Also, if I interpreted the data correctly, there are 241 values, not 240.
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