converting c for loop to fortran do loop - c++

Question

I am converting a cpp prog (from another author) to a Fortran prog, my C is not too strong. I came across for-loop constructs starting with
for (int n = 1; 1; ++n) {
...
I would have expected this to convert to a Fortran Do as per
Do n=1, 1, 2
...
... at least that is my guess based on my understanding of what ++n will do.
Is my translation correct? If so, the loop will cycle at most once, so what am I missing ???
I understand that in some ways c for-loops have a "do-while" aspect, and hence wrinkles porting to Fortran Do's.
Anyway ... a clarification would be much appreciated.
EDITED: after some prompt responses, and I think I see where this is going
First, the exact C code copy/paste but "trimming" a little, is
for (int n = 1; 1; ++n) {
const double coef = exp(-a2*(n*n)) * expx2 / (a2*(n*n) + y*y);
prod2ax *= exp2ax;
prodm2ax *= expm2ax;
sum1 += coef;
sum2 += coef * prodm2ax;
sum4 += (coef * prodm2ax) * (a*n);
sum3 += coef * prod2ax;
sum5 += (coef * prod2ax) * (a*n);
// test convergence via sum5, since this sum has the slowest decay
if ((coef * prod2ax) * (a*n) < relerr * sum5) break;
}
So yes, there is a "break" in the loop, which on the Fortran side is replaced with an "Exit".
I think the key seems to be from the answers below that the original code's author created the
for (int n=1; 1 ; ++n )
precisely to create a an infinite loop, and I had not guessed that this for construct would create an infinite loop.
Anyway, I can certainly create an infinite loop with an "Exit" in Fortran (though I expect I might "do" it a bit more judiciously)
Many thanks to all.
It seems the Mr Gregory's response was the one that imediately lead to a solution for me, so I will mark his correct. As for the Fortran side, there are a number of alternatives such as:
Do While
:
If( something ) Exit
End Do
but being old fashioned I would probably use a construct with a "limit" such as
Do i=1, MaxIter
:
If( something ) Exit
End Do
For slightly fancier applications I might include a return flag in case it did not converge in MaxIter's etc.

Answer 1

It's difficult to be definitive without seeing how the C++ program breaks out of that loop, but a straightforward Fortran equivalent would be
n = 1
do
! code, including an exit under some condition, presumably on the value of n
n = n+1
end do
If the loop is terminated when n reaches a critical value then the equivalent might be
do n = 1, critical_value ! no need to indicate step size if it is 1
! code
end do

Answer 2

Are you sure you wrote the C code correctly? Typically loops in C/C++ are done like this:
for (int n = 1; n < 10; ++n) {
// ...
}
Note the "n < 10" test condition. Your code's test condition is simply 1, which will always evaluate to Boolean "true". This means the code will loop infinitely, unless there's a break inside the loop, which you haven't shown.
++n means "increment n".
So if the code you've shown is indeed correct, the FORTRAN equivalent would be:
n = 1
do
[Body of the loop, which you haven't shown]
n = n + 1
enddo

Answer 3

Here's what
for (int n = 1; 1; ++n)
does:
It sets n to 1, then loops infinitely, incrementing n by 1 at the end of each loop iteration. The loop will never terminate unless something inside the loop breaks out.
It's been a long time since I wrote Fortran but as I recall the do loop you translated it to is not correct.

Answer 4

I don't think you can translate
for (int n = 1; 1; ++n)
to a FORTRAN DO loop. From what I recall, the notion of the generic conditional in C/C++ cannot be emulated in a FORTRAN DO loop.
The equivalent of
Do n=1, 1, 2
in C/C++ is
for ( int n = 1; n <= 1; n += 2 )

Answer 5

A few notes in addition to CareyGregory’s answer.
++n means ‘increment n by one (before n is evaluated)’
In C and C++, a for loop has three clauses, much like in FORTRAN:
for (init; condition; increment)
The difference is that each of the clauses must be a complete expression, whereas in FORTRAN the clauses are just values. It is just a ‘short’ way of writing an equivalent while loop:
int n = 1; │ for (int n = 1; 1; ++n) │ n = 1
while (1) │ { │ do
{ │ ... │ ...
... │ } │ n = n + 1
++n; │ │ enddo
} │ │