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*
************************************************************************
*
SUBROUTINE DLAMC5( BETA, P, EMIN, IEEE, EMAX, RMAX )
*
* -- LAPACK auxiliary routine (version 2.0) --
* Univ. of Tennessee, Univ. of California Berkeley, NAG Ltd.,
* Courant Institute, Argonne National Lab, and Rice University
* October 31, 1992
*
* .. Scalar Arguments ..
LOGICAL IEEE
INTEGER BETA, EMAX, EMIN, P
DOUBLE PRECISION RMAX
* ..
*
* Purpose
* =======
*
* DLAMC5 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.
*
* Arguments
* =========
*
* BETA (input) INTEGER
* The base of floating-point arithmetic.
*
* P (input) INTEGER
* The number of base BETA digits in the mantissa of a
* floating-point value.
*
* EMIN (input) INTEGER
* The minimum exponent before (gradual) underflow.
*
* IEEE (input) LOGICAL
* A logical flag specifying whether or not the arithmetic
* system is thought to comply with the IEEE standard.
*
* EMAX (output) INTEGER
* The largest exponent before overflow
*
* RMAX (output) DOUBLE PRECISION
* The largest machine floating-point number.
*
* =====================================================================
*
* .. Parameters ..
DOUBLE PRECISION ZERO, ONE
PARAMETER ( ZERO = 0.0D0, ONE = 1.0D0 )
* ..
* .. Local Scalars ..
INTEGER EXBITS, EXPSUM, I, LEXP, NBITS, TRY, UEXP
DOUBLE PRECISION OLDY, RECBAS, Y, Z
* ..
* .. External Functions ..
DOUBLE PRECISION DLAMC3
EXTERNAL DLAMC3
* ..
* .. 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 = DLAMC3( Y, Z )
20 CONTINUE
IF( Y.GE.ONE )
$ Y = OLDY
*
* Now multiply by BETA**EMAX to get RMAX.
*
DO 30 I = 1, EMAX
Y = DLAMC3( Y*BETA, ZERO )
30 CONTINUE
*
RMAX = Y
RETURN
*
* End of DLAMC5
*
END
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