Example 10. Constructing a polynomial expression.

The following constructs an expression representing a polynomial in $X$ with an arbitrary number of numerical coefficients:

*  Declare variables.
      INTEGER NCOEFF, STATUS, I, NSUBS
      REAL COEFF( NCOEFF )
      CHARACTER * ( * ) EXPRS

*  Initialise the expression.
      EXPRS = '<next_term><coeff>'                     [1]

*  Substitute each coefficient value.
      DO 1 I = 1, NCOEFF
         CALL TRN_STOKR( 'coeff', COEFF( I ), EXPRS,   [2]
     :                   NSUBS, STATUS )

*  Expand the token representing the next term.
         IF( I .NE. NCOEFF ) THEN
            CALL TRN_STOK( 'next_term',                [3]
     :                     '(<next_term><coeff>)*X+',
     :                     EXPRS, NSUBS, STATUS )

*  Eliminate the final '<next_term>' token.
         ELSE
            CALL TRN_STOK( 'next_term', ' ',           [4]
     :                     EXPRS, NSUBS, STATUS )
         ENDIF
   1  CONTINUE

Programming notes:

1. The character variable EXPRS, which is to contain the expression, is initialised to the value `<next_term><coeff>'.

2. The `<coeff>' token is substituted with a coefficient value (say 0.1) to give `<next_term>0.1'.

3. The `<next_term>' token is expanded to give `(<next_term><coeff>)*X+0.1'. The process is then repeated to replace the `<next_term><coeff>' part of this expression using the value of the next coefficient.

4. After the last coefficient value has been substituted the remaining `<next_term>' token is eliminated by replacing it with a blank string.

The effect of this algorithm with (say) the 5 polynomial coefficients 0.1, 0.12, 1.06, $-$4.4 & 1E$-$7, would be to produce the following expression:

'((((1E-7)*X+(-4.4))*X+1.06)*X+0.12)*X+0.1'

which casts the polynomial into a form for efficient evaluation using Horner's method.