slamc5()

subroutine slamc5	(	integer	BETA,
		integer	P,
		integer	EMIN,
		logical	IEEE,
		integer	EMAX,
		real	RMAX
	)

SLAMC5

Purpose:

 SLAMC5 attempts to compute RMAX, the largest machine floating-point
 number, without overflow.  It assumes that EMAX + abs(EMIN) sum
 approximately to a power of 2.  It will fail on machines where this
 assumption does not hold, for example, the Cyber 205 (EMIN = -28625,
 EMAX = 28718).  It will also fail if the value supplied for EMIN is
 too large (i.e. too close to zero), probably with overflow.

Parameters

[in]	BETA	The base of floating-point arithmetic.
[in]	P	The number of base BETA digits in the mantissa of a floating-point value.
[in]	EMIN	The minimum exponent before (gradual) underflow.
[in]	IEEE	A logical flag specifying whether or not the arithmetic system is thought to comply with the IEEE standard.
[out]	EMAX	The largest exponent before overflow
[out]	RMAX	The largest machine floating-point number.

Definition at line 802 of file slamchf77.f.

*
*  -- LAPACK auxiliary routine (version 3.7.0) --
*     Univ. of Tennessee, Univ. of California Berkeley and NAG Ltd..
*     November 2010
*
*     .. Scalar Arguments ..
      LOGICAL            IEEE
      INTEGER            BETA, EMAX, EMIN, P
      REAL               RMAX
*     ..
* =====================================================================
*
*     .. Parameters ..
      REAL               ZERO, ONE
      parameter( zero = 0.0e0, one = 1.0e0 )
*     ..
*     .. Local Scalars ..
      INTEGER            EXBITS, EXPSUM, I, LEXP, NBITS, TRY, UEXP
      REAL               OLDY, RECBAS, Y, Z
*     ..
*     .. External Functions ..
      REAL               SLAMC3
      EXTERNAL           slamc3
*     ..
*     .. Intrinsic Functions ..
      INTRINSIC          mod
*     ..
*     .. Executable Statements ..
*
*     First compute LEXP and UEXP, two powers of 2 that bound
*     abs(EMIN). We then assume that EMAX + abs(EMIN) will sum
*     approximately to the bound that is closest to abs(EMIN).
*     (EMAX is the exponent of the required number RMAX).
*
      lexp = 1
      exbits = 1
   10 CONTINUE
      try = lexp*2
      IF( try.LE.( -emin ) ) THEN
         lexp = try
         exbits = exbits + 1
         GO TO 10
      END IF
      IF( lexp.EQ.-emin ) THEN
         uexp = lexp
      ELSE
         uexp = try
         exbits = exbits + 1
      END IF
*
*     Now -LEXP is less than or equal to EMIN, and -UEXP is greater
*     than or equal to EMIN. EXBITS is the number of bits needed to
*     store the exponent.
*
      IF( ( uexp+emin ).GT.( -lexp-emin ) ) THEN
         expsum = 2*lexp
      ELSE
         expsum = 2*uexp
      END IF
*
*     EXPSUM is the exponent range, approximately equal to
*     EMAX - EMIN + 1 .
*
      emax = expsum + emin - 1
      nbits = 1 + exbits + p
*
*     NBITS is the total number of bits needed to store a
*     floating-point number.
*
      IF( ( mod( nbits, 2 ).EQ.1 ) .AND. ( beta.EQ.2 ) ) THEN
*
*        Either there are an odd number of bits used to store a
*        floating-point number, which is unlikely, or some bits are
*        not used in the representation of numbers, which is possible,
*        (e.g. Cray machines) or the mantissa has an implicit bit,
*        (e.g. IEEE machines, Dec Vax machines), which is perhaps the
*        most likely. We have to assume the last alternative.
*        If this is true, then we need to reduce EMAX by one because
*        there must be some way of representing zero in an implicit-bit
*        system. On machines like Cray, we are reducing EMAX by one
*        unnecessarily.
*
         emax = emax - 1
      END IF
*
      IF( ieee ) THEN
*
*        Assume we are on an IEEE machine which reserves one exponent
*        for infinity and NaN.
*
         emax = emax - 1
      END IF
*
*     Now create RMAX, the largest machine number, which should
*     be equal to (1.0 - BETA**(-P)) * BETA**EMAX .
*
*     First compute 1.0 - BETA**(-P), being careful that the
*     result is less than 1.0 .
*
      recbas = one / beta
      z = beta - one
      y = zero
      DO 20 i = 1, p
         z = z*recbas
         IF( y.LT.one )
     $      oldy = y
         y = slamc3( y, z )
   20 CONTINUE
      IF( y.GE.one )
     $   y = oldy
*
*     Now multiply by BETA**EMAX to get RMAX.
*
      DO 30 i = 1, emax
         y = slamc3( y*beta, zero )
   30 CONTINUE
*
      rmax = y
      RETURN
*
*     End of SLAMC5
*

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