Cloned SEACAS for EXODUS library with extra build files for internal package management.
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C Copyright(C) 1999-2020 National Technology & Engineering Solutions
C of Sandia, LLC (NTESS). Under the terms of Contract DE-NA0003525 with
C NTESS, the U.S. Government retains certain rights in this software.
C
C See packages/seacas/LICENSE for details
*++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
*
* symmlqblas fortran
*
* daxpy dcopy ddot dnrm2
*
*** from netlib, Thu May 16 21:00:13 EDT 1991 ***
*** Declarations of the form dx(1) changed to dx(*)
*
*++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
*++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
subroutine daxpy(n,da,dx,incx,dy,incy)
c constant times a vector plus a vector.
c uses unrolled loops for increments equal to one.
c jack dongarra, linpack, 3/11/78.
double precision dx(*),dy(*),da
integer i,incx,incy,ix,iy,m,mp1,n
if(n.le.0)return
if (da .eq. 0.0d0) return
if(incx.eq.1.and.incy.eq.1)go to 20
c code for unequal increments or equal increments
c not equal to 1
ix = 1
iy = 1
if(incx.lt.0)ix = (-n+1)*incx + 1
if(incy.lt.0)iy = (-n+1)*incy + 1
do 10 i = 1,n
dy(iy) = dy(iy) + da*dx(ix)
ix = ix + incx
iy = iy + incy
10 continue
return
c code for both increments equal to 1
c clean-up loop
20 m = mod(n,4)
if( m .eq. 0 ) go to 40
do 30 i = 1,m
dy(i) = dy(i) + da*dx(i)
30 continue
if( n .lt. 4 ) return
40 mp1 = m + 1
do 50 i = mp1,n,4
dy(i) = dy(i) + da*dx(i)
dy(i + 1) = dy(i + 1) + da*dx(i + 1)
dy(i + 2) = dy(i + 2) + da*dx(i + 2)
dy(i + 3) = dy(i + 3) + da*dx(i + 3)
50 continue
return
end
*++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
subroutine dcopy(n,dx,incx,dy,incy)
c copies a vector, x, to a vector, y.
c uses unrolled loops for increments equal to one.
c jack dongarra, linpack, 3/11/78.
double precision dx(*),dy(*)
integer i,incx,incy,ix,iy,m,mp1,n
if(n.le.0)return
if(incx.eq.1.and.incy.eq.1)go to 20
c code for unequal increments or equal increments
c not equal to 1
ix = 1
iy = 1
if(incx.lt.0)ix = (-n+1)*incx + 1
if(incy.lt.0)iy = (-n+1)*incy + 1
do 10 i = 1,n
dy(iy) = dx(ix)
ix = ix + incx
iy = iy + incy
10 continue
return
c code for both increments equal to 1
c clean-up loop
20 m = mod(n,7)
if( m .eq. 0 ) go to 40
do 30 i = 1,m
dy(i) = dx(i)
30 continue
if( n .lt. 7 ) return
40 mp1 = m + 1
do 50 i = mp1,n,7
dy(i) = dx(i)
dy(i + 1) = dx(i + 1)
dy(i + 2) = dx(i + 2)
dy(i + 3) = dx(i + 3)
dy(i + 4) = dx(i + 4)
dy(i + 5) = dx(i + 5)
dy(i + 6) = dx(i + 6)
50 continue
return
end
*++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
double precision function ddot(n,dx,incx,dy,incy)
c forms the dot product of two vectors.
c uses unrolled loops for increments equal to one.
c jack dongarra, linpack, 3/11/78.
double precision dx(*),dy(*),dtemp
integer i,incx,incy,ix,iy,m,mp1,n
ddot = 0.0d0
dtemp = 0.0d0
if(n.le.0)return
if(incx.eq.1.and.incy.eq.1)go to 20
c code for unequal increments or equal increments
c not equal to 1
ix = 1
iy = 1
if(incx.lt.0)ix = (-n+1)*incx + 1
if(incy.lt.0)iy = (-n+1)*incy + 1
do 10 i = 1,n
dtemp = dtemp + dx(ix)*dy(iy)
ix = ix + incx
iy = iy + incy
10 continue
ddot = dtemp
return
c code for both increments equal to 1
c clean-up loop
20 m = mod(n,5)
if( m .eq. 0 ) go to 40
do 30 i = 1,m
dtemp = dtemp + dx(i)*dy(i)
30 continue
if( n .lt. 5 ) go to 60
40 mp1 = m + 1
do 50 i = mp1,n,5
dtemp = dtemp + dx(i)*dy(i) + dx(i + 1)*dy(i + 1) +
* dx(i + 2)*dy(i + 2) + dx(i + 3)*dy(i + 3) + dx(i + 4)*dy(i + 4)
50 continue
60 ddot = dtemp
return
end
*++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
double precision function dnrm2 ( n, dx, incx)
integer next
double precision dx(*), cutlo, cuthi, hitest, sum, xmax,zero,one
data zero, one /0.0d0, 1.0d0/
c euclidean norm of the n-vector stored in dx() with storage
c increment incx .
c if n .le. 0 return with result = 0.
c if n .ge. 1 then incx must be .ge. 1
c c.l.lawson, 1978 jan 08
c four phase method using two built-in constants that are
c hopefully applicable to all machines.
c cutlo = maximum of dsqrt(u/eps) over all known machines.
c cuthi = minimum of dsqrt(v) over all known machines.
c where
c eps = smallest no. such that eps + 1. .gt. 1.
c u = smallest positive no. (underflow limit)
c v = largest no. (overflow limit)
c brief outline of algorithm..
c phase 1 scans zero components.
c move to phase 2 when a component is nonzero and .le. cutlo
c move to phase 3 when a component is .gt. cutlo
c move to phase 4 when a component is .ge. cuthi/m
c where m = n for x() real and m = 2*n for complex.
c values for cutlo and cuthi..
c from the environmental parameters listed in the imsl converter
c document the limiting values are as follows..
c cutlo, s.p. u/eps = 2**(-102) for honeywell. close seconds are
c univac and dec at 2**(-103)
c thus cutlo = 2**(-51) = 4.44089e-16
c cuthi, s.p. v = 2**127 for univac, honeywell, and dec.
c thus cuthi = 2**(63.5) = 1.30438e19
c cutlo, d.p. u/eps = 2**(-67) for honeywell and dec.
c thus cutlo = 2**(-33.5) = 8.23181d-11
c cuthi, d.p. same as s.p. cuthi = 1.30438d19
c data cutlo, cuthi / 8.232d-11, 1.304d19 /
c data cutlo, cuthi / 4.441e-16, 1.304e19 /
data cutlo, cuthi / 8.232d-11, 1.304d19 /
if(n .gt. 0) go to 10
dnrm2 = zero
go to 300
10 assign 30 to next
sum = zero
nn = n * incx
c begin main loop
i = 1
20 go to next,(30, 50, 70, 110)
30 if( dabs(dx(i)) .gt. cutlo) go to 85
assign 50 to next
xmax = zero
c phase 1. sum is zero
50 if( dx(i) .eq. zero) go to 200
if( dabs(dx(i)) .gt. cutlo) go to 85
c prepare for phase 2.
assign 70 to next
go to 105
c prepare for phase 4.
100 i = j
assign 110 to next
sum = (sum / dx(i)) / dx(i)
105 xmax = dabs(dx(i))
go to 115
c phase 2. sum is small.
c scale to avoid destructive underflow.
70 if( dabs(dx(i)) .gt. cutlo ) go to 75
c common code for phases 2 and 4.
c in phase 4 sum is large. scale to avoid overflow.
110 if( dabs(dx(i)) .le. xmax ) go to 115
sum = one + sum * (xmax / dx(i))**2
xmax = dabs(dx(i))
go to 200
115 sum = sum + (dx(i)/xmax)**2
go to 200
c prepare for phase 3.
75 sum = (sum * xmax) * xmax
c for real or d.p. set hitest = cuthi/n
c for complex set hitest = cuthi/(2*n)
85 hitest = cuthi/float( n )
c phase 3. sum is mid-range. no scaling.
do 95 j =i,nn,incx
if(dabs(dx(j)) .ge. hitest) go to 100
95 sum = sum + dx(j)**2
dnrm2 = dsqrt( sum )
go to 300
200 continue
i = i + incx
if ( i .le. nn ) go to 20
c end of main loop.
c compute square root and adjust for scaling.
dnrm2 = xmax * dsqrt(sum)
300 continue
return
end