real function snrm2 ( n, sx, incx) integer next real sx(1), cutlo, cuthi, hitest, sum, xmax, zero, one data zero, one /0.0e0, 1.0e0/ c c euclidean norm of the n-vector stored in sx() 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 c.l.lawson, 1978 jan 08 c c four phase method using two built-in constants that are c hopefully applicable to all machines. c cutlo = maximum of sqrt(u/eps) over all known machines. c cuthi = minimum of sqrt(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 c brief outline of algorithm.. c 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 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 / 4.441e-16, 1.304e19 / c if(n .gt. 0) go to 10 snrm2 = zero go to 300 c 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( abs(sx(i)) .gt. cutlo) go to 85 assign 50 to next xmax = zero c c phase 1. sum is zero c 50 if( sx(i) .eq. zero) go to 200 if( abs(sx(i)) .gt. cutlo) go to 85 c c prepare for phase 2. assign 70 to next go to 105 c c prepare for phase 4. c 100 i = j assign 110 to next sum = (sum / sx(i)) / sx(i) 105 xmax = abs(sx(i)) go to 115 c c phase 2. sum is small. c scale to avoid destructive underflow. c 70 if( abs(sx(i)) .gt. cutlo ) go to 75 c c common code for phases 2 and 4. c in phase 4 sum is large. scale to avoid overflow. c 110 if( abs(sx(i)) .le. xmax ) go to 115 sum = one + sum * (xmax / sx(i))**2 xmax = abs(sx(i)) go to 200 c 115 sum = sum + (sx(i)/xmax)**2 go to 200 c c c prepare for phase 3. c 75 sum = (sum * xmax) * xmax c c c for real or d.p. set hitest = cuthi/n c for complex set hitest = cuthi/(2*n) c 85 hitest = cuthi/float( n ) c c phase 3. sum is mid-range. no scaling. c do 95 j =i,nn,incx if(abs(sx(j)) .ge. hitest) go to 100 95 sum = sum + sx(j)**2 snrm2 = sqrt( sum ) go to 300 c 200 continue i = i + incx if ( i .le. nn ) go to 20 c c end of main loop. c c compute square root and adjust for scaling. c snrm2 = xmax * sqrt(sum) 300 continue return end