问题描述
我正在比较两个文件，并使用Fortran查找共处的点。代码在使用gfortran 10.2.0版的笔记本电脑上可以很好地工作，而我将此代码保存在ssh服务器上（ssh使用gfortran 4.8.5；也曾在8.3.1版本上尝试过），它在大多数迭代中都很好，但提供了几次迭代时出现以下错误：
Program received signal SIGSEGV: Segmentation fault - invalid memory reference.
Backtrace for this error:
#0 0x7F750B7F8697
#1 0x7F750B7F8CDE
#2 0x7F750ACF33FF
#3 0x402493 in MAIN__ at 1_myd_trmm_coloc.f90:?
1a_myd_trmm_coloc.sh: line 10: 42668 Segmentation fault (core dumped) ./1_myd_trmm_coloc.out < infile
以下是我的Fortran代码：
IMPLICIT NONE
Character*180::mydf,trmmf,outf
integer::my,mmn,md,mh,mmi,ms,mat
real::mla,mlo,ma412,ma470,ma550,ma660,mae
integer::ty,tmn,td,th,ic,i,j,i1,j1,i22,j22,n
real::tla,tlo,tpr
real,allocatable::m(:,:),t(:,:)
real::delat,delon,c,a,dist,rlon1,rlon2,rlat1,rlat2
real::mtim,ttim,maxttim,minttim,minlat,maxlat,minlon,maxlon
real,dimension(100000)::ds,ala,alo
integer,dimension(100000)::ay,amn,ad,ah,ami,as,aat
real,dimension(100000)::a412,a470,a550,a660,aae,ppr
read*,mydf
read*,trmmf
read*,outf
open(11,file=mydf,status='old')
i=0
5 read(11,*,end=90,err=80)my,mla,&
ma412,mae,mat
i=i+1
80 goto 5
90 close(11)
allocate(m(i,14))
open(11,status='old')
i1=0
6 read(11,end=91,err=81)my,&
ma412,mat
i1=i1+1
m(i1,1)=my ; m(i1,2)=mmn ; m(i1,3)=md
m(i1,4)=mh ; m(i1,5)=mmi ; m(i1,6)=ms
m(i1,7)=mla ; m(i1,8)=mlo ; m(i1,9)=ma412
m(i1,10)=ma470 ; m(i1,11)=ma550 ; m(i1,12)=ma660
m(i1,13)=mae ; m(i1,14)=mat
81 goto 6
91 close(11)
open(12,file=trmmf,status='old')
j=0
15 read(12,end=190,err=180)ty,tla,tpr
j=j+1
180 go to 15
190 close(12)
allocate(t(j,7))
open(12,status='old')
j1=0
16 read(12,end=191,err=181)ty,tpr
j1=j1+1
t(j1,1)=ty ; t(j1,2)=tmn ; t(j1,3)=td
t(j1,4)=th ; t(j1,5)=tla ; t(j1,6)=tlo
t(j1,7)=tpr ;
181 go to 16
191 close(12)
open(51,file=outf,status='unkNown')
do i22=1,i
n=0; ds=-999.0
ala=-999.0;alo=-999.0;a412=-999.0;a470=-999.0
a550=-999.0;a660=-999.0;aae=-999.0;aat=-999;
ppr=-999.0;
ic=0
do j22=1,j
if(t(j22,1).eq.m(i22,1).and.t(j22,2).eq.m(i22,2).and. &
t(j22,3).eq.m(i22,3)) then
minlat=t(j22,5)-0.125; maxlat=t(j22,5)+0.125
minlon=t(j22,6)-0.125; maxlon=t(j22,6)+0.125
mtim=m(j22,4)+(m(j22,5)/60)+(m(j22,6)/3600)
ttim=t(j22,4)
minttim=ttim-0.5; maxttim=ttim+0.5
if(m(i22,7).ge.minlat.and.m(i22,7).le.maxlat.and. &
m(i22,8).ge.minlon.and.m(i22,8).le.maxlon.and. &
mtim.ge.minttim.and.mtim.le.maxttim) then
rlat1=(m(i22,7)*3.141592657)/180.0; rlat2=(t(j22,5)*3.141592657)/180.0
rlon1=(m(i22,8)*3.141592657)/180.0; rlon2=(t(j22,6)*3.141592657)/180.0
delat=rlat2-rlat1; delon=rlon2-rlon1
a=SIN(delat/2.0)**2+(COS(rlat1)*COS(rlat2)*SIN(delon/2.0)**2)
c=2.0*atan2(sqrt(a),sqrt(1-a))
dist=6378136.300*c
ic=ic+1
! print*,ic
ds(ic)=dist
ala(ic)=m(i22,7)
alo(ic)=m(i22,8)
ay(ic)=m(i22,1); amn(ic)=m(i22,2); ad(ic)=m(i22,3)
ah(ic)=m(i22,4); ami(ic)=m(i22,5); as(ic)=m(i22,6)
a412(ic)=m(i22,9); a470(ic)=m(i22,10)
a550(ic)=m(i22,11); a660(ic)=m(i22,12)
aae(ic)=m(i22,13); aat(ic)=m(i22,14)
! py(ic)=t(j22,1); pmn(ic)=t(j22,2); pd(ic)=t(j22,3)
! ph(ic)=t(j22,4); pla(ic)=t(j22,5); plo(ic)=t(j22,6)
ppr(ic)=t(j22,7)
end if
end if
end do
if (ic>0)then
write(51,111)minval(ds,ds.ne.-999.0),ay(minloc(ds,ds.ne.-999.0)),&
amn(minloc(ds,ad(minloc(ds,&
ah(minloc(ds,ami(minloc(ds,&
as(minloc(ds,ala(minloc(ds,&
alo(minloc(ds,a412(minloc(ds,&
a470(minloc(ds,a550(minloc(ds,&
a660(minloc(ds,aae(minloc(ds,&
aat(minloc(ds,ds.ne.-999)),ppr(minloc(ds,ds.ne.-999.0))
n=n+1
! print*,n
end if
end do
111 format(f16.6,1x,i4,5(1x,i2.2),2(1x,f8.3),f6.3),i2,f8.3)
deallocate(m)
deallocate(t)
close(51)
end
文件1
2008 04 06 09 27 03 10.250 50.250 0.303 0.303 0.303 0.303 0.000 1
2008 04 06 09 27 03 10.500 50.500 0.247 0.247 0.247 0.247 0.000 1
2008 04 06 09 27 03 10.750 50.500 0.219 0.180 0.142 0.111 1.500 2
2008 04 06 09 27 03 11.000 50.500 0.157 0.129 0.102 0.079 1.500 2
2008 04 06 09 27 46 11.250 50.000 0.064 0.053 0.042 0.032 1.500 2
2008 04 06 09 27 46 11.250 50.250 0.151 0.124 0.098 0.076 1.500 2
2008 04 06 09 27 46 11.500 50.250 0.129 0.106 0.083 0.065 1.500 2
文件2
2008 4 6 6 9.875 29.125 0
2008 4 6 6 10.125 29.125 0
2008 4 6 6 10.375 29.125 0
2008 4 6 6 10.625 29.125 0
2008 4 6 6 10.875 29.125 0
2008 4 6 6 11.125 29.125 0
2008 4 6 6 11.375 29.125 0
2008 4 6 6 11.625 29.125 0
2008 4 6 6 11.875 29.125 0
2008 4 6 6 12.125 29.125 0
解决方法
我没有可以测试的版本，但我怀疑您的问题出自aat（minloc（ds，ds.ne.-999））-您正在针对-999的整数测试实际ds，这很少好主意。此外（仅供参考，仅在少数情况下才能测试实数之间的等式。
注意发布样本数据集将有助于进一步诊断