Problem in Fortran code

[ISAIAH]

Hello,
I have done a program which solves each iteration of a, but it is giving me an error, here is my source code, please check it.

real k, z, y (500), x, Da, Q, v, a
k = 0.0005
Q = 0.026
v = 52
z = 0.0048
x = 0.003
Da = 0.2
do
a = 0.41
y (a) = y (a) + ((k * (z-(2 * y)) * (x-y) * Da) / (v + (Q * a)))
enddo
if (a.le.45) ahen
a = a +0.1
endif
wriae (*,*) y (a)
end

[kelfro]

Hello,

real k, z, y (500), x, Da, Q, v, a
...
a = 0.41
...

Well, I think this is a recognition of Pascal. "a" seem to be real, but it is assigned with a comma, should scream. What happens if you assign-est by a = 0.41?

[opaper]

Hi,

Well, I think this is a recognition of Pascal.

No, in Pascal, we would have written

... k, z, y (500), x, Dt, Q, v, a: real

as well as the type real of Pascal, which is actually from Borland's Turbo, Pascal is not recommended to use in a specific type, 6-byte, and for which the compiler does not use the FPU, but only with the CPU, and the injury time calculations.

"a" seem to be real, but it is assigned with a comma, should scream.

Unless the compiler is smart enough to use representations which are via localization.

[ISAIAH]

Hello
Thank you for your reply.

as well as the type real of Pascal, which is actually from Borland's Turbo, Pascal is not recommended to use in a specific type, 6-byte, and for which the compiler does not use the FPU, but only with the CPU, and the injury time calculations.

So now, if I take a = 0.41, it will consider that the value of a is 0.41. But I think a varies from 0 to 41. Am I correct? Please guide me with this. I think I have not understood the concept, so please give me some suggestion on my code.

[Modifier]

Hi,
I find some of the problems, first

do
a = 0.45
y (a) = y (a) + ((k * (z-(2 * y)) * (x-y) * Da) / (v + (Q * a)))
enddo

As I understand you intend to loop a, a is ranging from 0 to 45. In this case, I recommend you to write a single line.

do a = 0.41

In this case, T must be declared as an integer. If a is actually real, you have a loop "infinite", of which you will never leave if you do not add a condition "exit" somewhere between "do" and "enddo". Moreover, if t is real, it must be initialized as,

a = 0.45

You should not use a comma, as reported before.

[Reegan]

Hello,
Correct the above mentioned. Other problems are, you iterate over the array y. Let a be an integer, or it's real. In the latter case, the assignment to a real value will cause big problems with its use as an index of an array.
Another problem: you use y (a) to define y (a). Now, y (a) is not initialized, it can take a zero value at best and at worst based on any compiler that you use.
Subsequently:

if (a.le.45) then
a = a +0.1
endif

Again, in writing the condition of "if", we understand that a is an integer, otherwise you would rather something like

if (a.le.45.)

However, just after you add 0.1 to a, as if it was real.
Finally:

write (*,*) y (a)

He will write to the screen the value of y (a), for actual value of a it knows, Since a is real, it will not know how to make an array index. If you want to write the entire table, you must either pass through a loop "enddo do ...", you simply write

write (*,*) y

but it will write the entire table (500 numbers), which are obviously not all interesting. The most delicate, in my opinion, is to disambiguate between a and a real world, using two different variables.