Name: wen-long sang (email_not_shown)
Date: 02/17/16-06:49:49 AM Z


Dear Vladyslav,

   Thank you for your reply! You are right, "D" corresponds to double precision in fortran. Actually, I need a double precision expression in fortran. However, abc = 0.6*(-8 + 12 Log[5 + y[1]]) and abc1 = 1/x^(2/3) will be translated to abc = 6.D-1*(-8D0 + 12D1*Log(5D0 + y(1D0))) and abc1 = x**-6.666666666666666D0D-1, where 12 ==> 12D1 (12D1=120), and -2/3 ==> -6.666666666666666D0D-1 (fortran may not understand D0D-1). So is this a bug?
 
  Best Regards
 Wen-Long



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