Fortran 77代码可以工作,但不能显示所有值
我的程序可以运行,但它不能显示所有的p(I)值 当其工作时,仅显示p(1)值。停下来。但我想知道所有的价值观。P(1)至P(121) 问题出在哪里?循环是否有问题?Fortran 77代码可以工作,但不能显示所有值,fortran,fortran77,Fortran,Fortran77,我的程序可以运行,但它不能显示所有的p(I)值 当其工作时,仅显示p(1)值。停下来。但我想知道所有的价值观。P(1)至P(121) 问题出在哪里?循环是否有问题? PROGRAM odev dimension P(121) c div(ro.v.fi)=div(r.gradfi)+a-bfi c P(1)=P(2)=P(3)=P(4)=P(5)=P(6)=P(7)=P(8)=P(9)=P(10)=P(11)=100 c P(12
PROGRAM odev
dimension P(121)
c div(ro.v.fi)=div(r.gradfi)+a-bfi
c P(1)=P(2)=P(3)=P(4)=P(5)=P(6)=P(7)=P(8)=P(9)=P(10)=P(11)=100
c P(12)=P(23)=P(34)=P(45)=P(56)=P(67)=P(78)=P(89)=P(100)=P(111)=100
c P(22)=P(33)=P(44)=P(55)=P(66)=P(77)=P(88)=P(99)=P(110)=P(121)=0
c P(112)=P(113)=P(114)=P(115)=P(116)=P(117)=P(118)=P(119)=P(120)=0
50 PRINT *, "Hangi yontemle cozum yapmak istiyorsunuz?"
PRINT *, "Merkezi farklar icin 1"
PRINT *, "Upwind icin 2"
PRINT *, "Hybrid icin 3"
PRINT *, "Powerlaw icin 4 giriniz"
READ *, DE
PRINT *, "iterasyon saysn giriniz"
read*, iter
IF (DE.eq.1) THEN
GO TO 10
ELSE IF (DE.eq.2) THEN
GO TO 20
ELSE IF (DE.eq.3) THEN
GO TO 30
ELSE IF (DE.eq.4) THEN
GO TO 40
ELSE
PRINT *, "Lutfen 1-4 aral§nda giriŸ yapnz"
GO TO 50
END IF
5 do I=1,11
P(I)=100
end do
6 do I=12,111,11
P(I)= 100
end do
7 do I=22,121,11
P(I)=0
end do
8 do I=112,120
P(I)=0
end do
10 PRINT *, "Merkezi farklar metodu"
c tanimlanan formule gore 10*10 hucreli grid
c sistem icin hesap yapar
do 3 n=1, iter
DO 4 I=13, 109
P(I)=(0.5*P(I+1)+1.5*P(I-1)-P(I-11)+3*P(I+11)+10)/6
4 continue
3 continue
GO TO 60
20 PRINT *, "Upwind metodu "
do n=1, iter
DO I=13, 109
P(I)=(P(I+1)+2*P(I-1)+P(I-11)+5*P(I+11)+10)/11
END DO
end do
GO TO 60
30 PRINT *, "Hybrid metodu "
do n=1, iter
DO I=13, 109
P(I)=(0.5*P(I+1)+1.5*P(I-1)+4*P(I+11)+10)/8
END DO
end do
GO TO 60
40 PRINT *, "Powerlaw metodu "
do n=1, iter
DO I=13, 109
P(I)=(0.591*P(I+1)+1.591*P(I-1)+0.078*P(I-11)+4.078*P(I+11)+10)/8.338
END DO
end do
GO TO 60
60 PRINT *, "Tesekkurler"
do 11 I=1,121
print *, I, P(I)
pause
11 continue
END
编辑
来自高绩效分数的建议
c P(112)=P(113)=P(114)=P(115)=P(116)=P(117)=P(118)=P(119)=P(120)=0
5 do I=1,11
P(I)=100
end do
6 do I=12,111,11
P(I)= 100
end do
7 do I=22,121,11
P(I)=0
end do
8 do I=112,120
P(I)=0
end do
50 PRINT *, "Hangi yontemle cozum yapmak istiyorsunuz?"
SELECT CASE (N)
CASE (1)
PRINT *, "Merkezi farklar metodu"
do 3 n=1, iter
DO 4 I=13, 109
P(I)=(0.5*P(I+1)+1.5*P(I-1)-P(I-11)+3*P(I+11)+10)/6
4 continue
3 continue
GO TO 60
CASE (2)
PRINT *, "Upwind metodu "
do n=1, iter
DO I=13, 109
P(I)=(P(I+1)+2*P(I-1)+P(I-11)+5*P(I+11)+10)/11
END DO
end do
GO TO 60
CASE (3)
PRINT *, "Hybrid metodu "
do n=1, iter
DO I=13, 109
P(I)=(0.5*P(I+1)+1.5*P(I-1)+4*P(I+11)+10)/8
END DO
end do
GO TO 60
CASE (4)
PRINT *, "Powerlaw metodu "
do n=1, iter
DO I=13, 109
Z=8.338
P(I)=(0.591*P(I+1)+1.591*P(I-1)+0.078*P(I-11)+4.078*P(I+11)+10)/Z
END DO
end do
GO TO 60
END SELECT
现在是工作。但所示的P(13)至P(109)值>Nan
p(1)至p(12)接近1e-40(必须为100)
P(110)到P(121)接近1e-38
我想还有一个问题
编辑2
来自@francescalus的建议
c P(112)=P(113)=P(114)=P(115)=P(116)=P(117)=P(118)=P(119)=P(120)=0
5 do I=1,11
P(I)=100
end do
6 do I=12,111,11
P(I)= 100
end do
7 do I=22,121,11
P(I)=0
end do
8 do I=112,120
P(I)=0
end do
50 PRINT *, "Hangi yontemle cozum yapmak istiyorsunuz?"
SELECT CASE (N)
CASE (1)
PRINT *, "Merkezi farklar metodu"
do 3 n=1, iter
DO 4 I=13, 109
P(I)=(0.5*P(I+1)+1.5*P(I-1)-P(I-11)+3*P(I+11)+10)/6
4 continue
3 continue
GO TO 60
CASE (2)
PRINT *, "Upwind metodu "
do n=1, iter
DO I=13, 109
P(I)=(P(I+1)+2*P(I-1)+P(I-11)+5*P(I+11)+10)/11
END DO
end do
GO TO 60
CASE (3)
PRINT *, "Hybrid metodu "
do n=1, iter
DO I=13, 109
P(I)=(0.5*P(I+1)+1.5*P(I-1)+4*P(I+11)+10)/8
END DO
end do
GO TO 60
CASE (4)
PRINT *, "Powerlaw metodu "
do n=1, iter
DO I=13, 109
Z=8.338
P(I)=(0.591*P(I+1)+1.591*P(I-1)+0.078*P(I-11)+4.078*P(I+11)+10)/Z
END DO
end do
GO TO 60
END SELECT
来自@agentp建议
c P(112)=P(113)=P(114)=P(115)=P(116)=P(117)=P(118)=P(119)=P(120)=0
5 do I=1,11
P(I)=100
end do
6 do I=12,111,11
P(I)= 100
end do
7 do I=22,121,11
P(I)=0
end do
8 do I=112,120
P(I)=0
end do
50 PRINT *, "Hangi yontemle cozum yapmak istiyorsunuz?"
SELECT CASE (N)
CASE (1)
PRINT *, "Merkezi farklar metodu"
do 3 n=1, iter
DO 4 I=13, 109
P(I)=(0.5*P(I+1)+1.5*P(I-1)-P(I-11)+3*P(I+11)+10)/6
4 continue
3 continue
GO TO 60
CASE (2)
PRINT *, "Upwind metodu "
do n=1, iter
DO I=13, 109
P(I)=(P(I+1)+2*P(I-1)+P(I-11)+5*P(I+11)+10)/11
END DO
end do
GO TO 60
CASE (3)
PRINT *, "Hybrid metodu "
do n=1, iter
DO I=13, 109
P(I)=(0.5*P(I+1)+1.5*P(I-1)+4*P(I+11)+10)/8
END DO
end do
GO TO 60
CASE (4)
PRINT *, "Powerlaw metodu "
do n=1, iter
DO I=13, 109
Z=8.338
P(I)=(0.591*P(I+1)+1.591*P(I-1)+0.078*P(I-11)+4.078*P(I+11)+10)/Z
END DO
end do
GO TO 60
END SELECT
现在终于可以工作了…这些行
print *, I, P(I)
pause
告诉程序打印p
的第一个元素,然后打印pause
。在许多计算机上,任何用户当时的输入,如按键,都会导致程序继续。在您的代码中,将导致P
的下一个元素写入终端,并且程序再次暂停
试着把这行代码暂停出来,看看会发生什么。或者像猴子一样坐在那里,按一个键120次
然后扔掉程序,用现代Fortran重写它;看FORTRAN77让我很痛苦。的初始赋值
p
(标记为5到10的行)从未执行过。无论DE
的值是什么,以及随后的跳转,P
在定义之前都会被引用。在这种情况下,获得想要的结果只是运气的问题。这是一个很好的例子,说明了为什么永远不应该使用goto
。。使用SELECT CASE
,(或if-then
结构,如果您真的必须使用f77)重写它,很高兴它能工作——您不需要SELECT CASE
中的goto
,但是@agentp已经观察到,每个案例末尾的goto
语句是多余的。它使代码看起来像C,令人不寒而栗。