dlamch.c

SuperLU 2.2版本。对大型、稀疏、非对称的线性系统的直接求解

/* +       WHILE( ( LEPS.GT.B ).AND.( B.GT.ZERO ) )LOOP */
L10:
	if (leps > b && b > zero) {
	    leps = b;
	    d__1 = half * leps;
/* Computing 5th power */
	    d__3 = two, d__4 = d__3, d__3 *= d__3;
/* Computing 2nd power */
	    d__5 = leps;
	    d__2 = d__4 * (d__3 * d__3) * (d__5 * d__5);
	    c = dlamc3_(&d__1, &d__2);
	    d__1 = -c;
	    c = dlamc3_(&half, &d__1);
	    b = dlamc3_(&half, &c);
	    d__1 = -b;
	    c = dlamc3_(&half, &d__1);
	    b = dlamc3_(&half, &c);
	    goto L10;
	}
/* +       END WHILE */

	if (a < leps) {
	    leps = a;
	}

/*        Computation of EPS complete.   

          Now find  EMIN.  Let A = + or - 1, and + or - (1 + BASE**(-3
)).   
          Keep dividing  A by BETA until (gradual) underflow occurs. T
his   
          is detected when we cannot recover the previous A. */

	rbase = one / lbeta;
	small = one;
	for (i = 1; i <= 3; ++i) {
	    d__1 = small * rbase;
	    small = dlamc3_(&d__1, &zero);
/* L20: */
	}
	a = dlamc3_(&one, &small);
	dlamc4_(&ngpmin, &one, &lbeta);
	d__1 = -one;
	dlamc4_(&ngnmin, &d__1, &lbeta);
	dlamc4_(&gpmin, &a, &lbeta);
	d__1 = -a;
	dlamc4_(&gnmin, &d__1, &lbeta);
	ieee = FALSE_;

	if (ngpmin == ngnmin && gpmin == gnmin) {
	    if (ngpmin == gpmin) {
		lemin = ngpmin;
/*            ( Non twos-complement machines, no gradual under
flow;   
                e.g.,  VAX ) */
	    } else if (gpmin - ngpmin == 3) {
		lemin = ngpmin - 1 + lt;
		ieee = TRUE_;
/*            ( Non twos-complement machines, with gradual und
erflow;   
                e.g., IEEE standard followers ) */
	    } else {
		lemin = min(ngpmin,gpmin);
/*            ( A guess; no known machine ) */
		iwarn = TRUE_;
	    }

	} else if (ngpmin == gpmin && ngnmin == gnmin) {
	    if ((i__1 = ngpmin - ngnmin, abs(i__1)) == 1) {
		lemin = max(ngpmin,ngnmin);
/*            ( Twos-complement machines, no gradual underflow
;   
                e.g., CYBER 205 ) */
	    } else {
		lemin = min(ngpmin,ngnmin);
/*            ( A guess; no known machine ) */
		iwarn = TRUE_;
	    }

	} else if ((i__1 = ngpmin - ngnmin, abs(i__1)) == 1 && gpmin == gnmin)
		 {
	    if (gpmin - min(ngpmin,ngnmin) == 3) {
		lemin = max(ngpmin,ngnmin) - 1 + lt;
/*            ( Twos-complement machines with gradual underflo
w;   
                no known machine ) */
	    } else {
		lemin = min(ngpmin,ngnmin);
/*            ( A guess; no known machine ) */
		iwarn = TRUE_;
	    }

	} else {
/* Computing MIN */
	    i__1 = min(ngpmin,ngnmin), i__1 = min(i__1,gpmin);
	    lemin = min(i__1,gnmin);
/*         ( A guess; no known machine ) */
	    iwarn = TRUE_;
	}
/* **   
   Comment out this if block if EMIN is ok */
	if (iwarn) {
	    first = TRUE_;
	    printf("\n\n WARNING. The value EMIN may be incorrect:- ");
	    printf("EMIN = %8i\n",lemin);
	    printf("If, after inspection, the value EMIN looks acceptable");
            printf("please comment out \n the IF block as marked within the"); 
            printf("code of routine DLAMC2, \n otherwise supply EMIN"); 
            printf("explicitly.\n");
	}
/* **   

          Assume IEEE arithmetic if we found denormalised  numbers abo
ve,   
          or if arithmetic seems to round in the  IEEE style,  determi
ned   
          in routine DLAMC1. A true IEEE machine should have both  thi
ngs   
          true; however, faulty machines may have one or the other. */

	ieee = ieee || lieee1;

/*        Compute  RMIN by successive division by  BETA. We could comp
ute   
          RMIN as BASE**( EMIN - 1 ),  but some machines underflow dur
ing   
          this computation. */

	lrmin = 1.;
	i__1 = 1 - lemin;
	for (i = 1; i <= 1-lemin; ++i) {
	    d__1 = lrmin * rbase;
	    lrmin = dlamc3_(&d__1, &zero);
/* L30: */
	}

/*        Finally, call DLAMC5 to compute EMAX and RMAX. */

	dlamc5_(&lbeta, &lt, &lemin, &ieee, &lemax, &lrmax);
    }

    *beta = lbeta;
    *t = lt;
    *rnd = lrnd;
    *eps = leps;
    *emin = lemin;
    *rmin = lrmin;
    *emax = lemax;
    *rmax = lrmax;

    return 0;


/*     End of DLAMC2 */

} /* dlamc2_ */


double dlamc3_(double *a, double *b)
{
/*  -- 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   


    Purpose   
    =======   

    DLAMC3  is intended to force  A  and  B  to be stored prior to doing 
  
    the addition of  A  and  B ,  for use in situations where optimizers 
  
    might hold one of these in a register.   

    Arguments   
    =========   

    A, B    (input) DOUBLE PRECISION   
            The values A and B.   

   ===================================================================== 
*/
/* >>Start of File<<   
       System generated locals */
    double ret_val;

    ret_val = *a + *b;

    return ret_val;

/*     End of DLAMC3 */

} /* dlamc3_ */


/* Subroutine */ int dlamc4_(int *emin, double *start, int *base)
{
/*  -- 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   


    Purpose   
    =======   

    DLAMC4 is a service routine for DLAMC2.   

    Arguments   
    =========   

    EMIN    (output) EMIN   
            The minimum exponent before (gradual) underflow, computed by 
  
            setting A = START and dividing by BASE until the previous A   
            can not be recovered.   

    START   (input) DOUBLE PRECISION   
            The starting point for determining EMIN.   

    BASE    (input) INT   
            The base of the machine.   

   ===================================================================== 
*/
    /* System generated locals */
    int i__1;
    double d__1;
    /* Local variables */
    static double zero, a;
    static int i;
    static double rbase, b1, b2, c1, c2, d1, d2;
    extern double dlamc3_(double *, double *);
    static double one;

    a = *start;
    one = 1.;
    rbase = one / *base;
    zero = 0.;
    *emin = 1;
    d__1 = a * rbase;
    b1 = dlamc3_(&d__1, &zero);
    c1 = a;
    c2 = a;
    d1 = a;
    d2 = a;
/* +    WHILE( ( C1.EQ.A ).AND.( C2.EQ.A ).AND.   
      $       ( D1.EQ.A ).AND.( D2.EQ.A )      )LOOP */
L10:
    if (c1 == a && c2 == a && d1 == a && d2 == a) {
	--(*emin);
	a = b1;
	d__1 = a / *base;
	b1 = dlamc3_(&d__1, &zero);
	d__1 = b1 * *base;
	c1 = dlamc3_(&d__1, &zero);
	d1 = zero;
	i__1 = *base;
	for (i = 1; i <= *base; ++i) {
	    d1 += b1;
/* L20: */
	}
	d__1 = a * rbase;
	b2 = dlamc3_(&d__1, &zero);
	d__1 = b2 / rbase;
	c2 = dlamc3_(&d__1, &zero);
	d2 = zero;
	i__1 = *base;
	for (i = 1; i <= *base; ++i) {
	    d2 += b2;
/* L30: */
	}
	goto L10;
    }
/* +    END WHILE */

    return 0;

/*     End of DLAMC4 */

} /* dlamc4_ */


/* Subroutine */ int dlamc5_(int *beta, int *p, int *emin, 
	int *ieee, int *emax, double *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   


    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) INT   
            The base of floating-point arithmetic.   

    P       (input) INT   
            The number of base BETA digits in the mantissa of a   
            floating-point value.   

    EMIN    (input) INT   
            The minimum exponent before (gradual) underflow.   

    IEEE    (input) INT   
            A int flag specifying whether or not the arithmetic   
            system is thought to comply with the IEEE standard.   

    EMAX    (output) INT   
            The largest exponent before overflow   

    RMAX    (output) DOUBLE PRECISION   
            The largest machine floating-point number.   

   ===================================================================== 
  


       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). */
    /* Table of constant values */
    static double c_b5 = 0.;
    
    /* System generated locals */
    int i__1;
    double d__1;
    /* Local variables */
    static int lexp;
    static double oldy;
    static int uexp, i;
    static double y, z;
    static int nbits;
    extern double dlamc3_(double *, double *);
    static double recbas;
    static int exbits, expsum, try__;



    lexp = 1;
    exbits = 1;
L10:
    try__ = lexp << 1;
    if (try__ <= -(*emin)) {
	lexp = try__;
	++exbits;
	goto L10;
    }
    if (lexp == -(*emin)) {
	uexp = lexp;
    } else {
	uexp = try__;
	++exbits;
    }

/*     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 > -lexp - *emin) {
	expsum = lexp << 1;
    } else {
	expsum = uexp << 1;
    }

/*     EXPSUM is the exponent range, approximately equal to   
       EMAX - EMIN + 1 . */

    *emax = expsum + *emin - 1;
    nbits = exbits + 1 + *p;

/*     NBITS is the total number of bits needed to store a   
       floating-point number. */

    if (nbits % 2 == 1 && *beta == 2) {

/*        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-b
it   
          system. On machines like Cray, we are reducing EMAX by one 
  
          unnecessarily. */

	--(*emax);
    }

    if (*ieee) {

/*        Assume we are on an IEEE machine which reserves one exponent
   
          for infinity and NaN. */

	--(*emax);
    }

/*     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 = 1. / *beta;
    z = *beta - 1.;
    y = 0.;
    i__1 = *p;
    for (i = 1; i <= *p; ++i) {
	z *= recbas;
	if (y < 1.) {
	    oldy = y;
	}
	y = dlamc3_(&y, &z);
/* L20: */
    }
    if (y >= 1.) {
	y = oldy;
    }

/*     Now multiply by BETA**EMAX to get RMAX. */

    i__1 = *emax;
    for (i = 1; i <= *emax; ++i) {
	d__1 = y * *beta;
	y = dlamc3_(&d__1, &c_b5);
/* L30: */
    }

    *rmax = y;
    return 0;

/*     End of DLAMC5 */

} /* dlamc5_ */

double pow_di(double *ap, int *bp)
{
    double pow, x;
    int n;

    pow = 1;
    x = *ap;
    n = *bp;

    if(n != 0){
	if(n < 0) {
	    n = -n;
	    x = 1/x;
	}
	for( ; ; ) {
	    if(n & 01) pow *= x;
	    if(n >>= 1)	x *= x;
	    else break;
	}
    }
    return(pow);
}