Printing REALs with the E Descriptor

WARNING: This example assumes the output is sent to a printer, and as a result, every formatted output contains printer control.

Problem Statement

Write a program that for x = 2.0, 1.8, 1.6, 1.4, ..., 0.8, 0.6, 0.4 and 0.2, prints the sequence number of the value (i.e., 1, 2, 3 and so on), the value itself, and its LOG(x) four times using E, E with three digits for the exponent, ES and EN descriptors. You should generate the output as shown below.

         1    1    2    2    3    3    4    4    5    5    6    6    7    7
....5....0....5....0....5....0....5....0....5....0....5....0....5....0....5
  1   2.00000    0.69315E+00   0.69315E+000    6.93147E-01  693.14718E-03
  2   1.80000    0.58779E+00   0.58779E+000    5.87787E-01  587.78661E-03
  3   1.60000    0.47000E+00   0.47000E+000    4.70004E-01  470.00358E-03
                                   :
                                   :
                                   :
 10   0.20000   -0.16094E+01  -0.16094E+001   -1.60944E+00   -1.60944E+00

Solution

PROGRAM  LogFunction
   IMPLICIT   NONE
   REAL, PARAMETER   :: Initial =  2.0
   REAL, PARAMETER   :: Final   =  0.0
   REAL, PARAMETER   :: Step    = -0.2
   REAL              :: x
   INTEGER           :: Count
   CHARACTER(LEN=80) :: FMT

   FMT   = "(I3, F10.5, E15.5, E15.5E3, ES15.5, EN15.5)"
   Count = 1
   x     = Initial
   DO
      IF (x <= Final)  EXIT
      WRITE(*,FMT)     Count, x, LOG(x), LOG(x), LOG(x), LOG(x)
      x     = x + Step
      Count = Count + 1
   END DO
END PROGRAM  LogFunction

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Program Input and Output

The output of the program is:

         1    1    2    2    3    3    4    4    5    5    6    6    7    7
....5....0....5....0....5....0....5....0....5....0....5....0....5....0....5
  1   2.00000    0.69315E+00   0.69315E+000    6.93147E-01  693.14718E-03
  2   1.80000    0.58779E+00   0.58779E+000    5.87787E-01  587.78661E-03
  3   1.60000    0.47000E+00   0.47000E+000    4.70004E-01  470.00358E-03
  4   1.40000    0.33647E+00   0.33647E+000    3.36472E-01  336.47212E-03
  5   1.20000    0.18232E+00   0.18232E+000    1.82321E-01  182.32140E-03
  6   1.00000   -0.17881E-06  -0.17881E-006   -1.78814E-07 -178.81395E-09
  7   0.80000   -0.22314E+00  -0.22314E+000   -2.23144E-01 -223.14376E-03
  8   0.60000   -0.51083E+00  -0.51083E+000   -5.10826E-01 -510.82587E-03
  9   0.40000   -0.91629E+00  -0.91629E+000   -9.16291E-01 -916.29112E-03
 10   0.20000   -0.16094E+01  -0.16094E+001   -1.60944E+00   -1.60944E+00

Discussion

Since from 2.0 down to 0.2 with a step size 0.2 has only 10 numbers, the sequence number has no more than 2 digits and can be printed with I3 with the first position being space.
For each of the remaining REAL values, we shall use five positions for the fractional part. Therefore, the value of x is printed F10.5. The four LOG(x) values are printed with E15.5, E15.5E3 with three positions for the exponent, ES15.5 and EN15.5. Hence, the format is
```
CHARACTER(LEN=80) :: FMT

FMT   = "(I3, F10.5, E15.5, E15.5E3, ES15.5, EN15.5)"
```
Since the last four values are the same but printed with different edit descriptors, you can compare their appearances. Of the last four values, the first and second are almost identical, except that the former has two positions for the exponent while the latter uses three. The third is printed with the ES descriptor, the integral part always has a non-zero digit and, as a result, the exponent value is one less than those using E. The last value uses EN, which causes Fortran to use no more than three digits in the integral part. For example, the value of LOG(2.0) (the first row) when printed with E15.5 is 0.69315E+00. If it is printed using EN15.5, the decimal point is shifted to the right three positions and the exponent is decreased by 3 generating an output 693.14718E-03. For the last value(i.e., 0.2), we have -0.16094E+01 using the E edit descriptor. If it is printed using the EN edit descriptor, the result is -1.60944E+00. Why? Recall that for the engineering notation, the exponent is a multiple of three and therefore, the possible values are 0, 3, -3, 6, -6, 9, -9, and so on. Since moving the decimal point in -0.16094E+01 to the right one position makes the exponent a zero, the printed result is -1.60944E+00.