# Won't Compile the simplest of programs! Am I being stupid???

by JamesWMH
Tags: c++, compile, fortran, program, programming
 P: 3 Hi, I'm just getting my teeth back into Fortran 90 I had a course on it last year, and just want to familirise myself with it before I apply for post grad stuff. Anyway, the program i have written is: program emosix real:: a, roota write(*,*) 'Type an interger to find the square root (must be a positive number)' read(*,*) a if(a>=0.0) stop roota = sqrt(a) write(*,*) 'The sqaure root of ', a, ' = ', roota pause end program emosix and I get the error code: Error: Cannot assign to a named constant at (1) C:\Users\James\Documents\Programming\Emo6\~emo6.f90:15.5: pause 1 Warning: Deleted feature: PAUSE statement at (1) The warning about deleted feature hasn't made any difference in the past, so I don't think its that, but I have no idea what the cannot assign... part means?? Any ideas? Thanks Jim
 Mentor P: 19,859 Why is that stop in there after the if statement? Also, your if statement needs an end if, I believe. program emosix real:: a, roota write(*,*) 'Type an interger to find the square root (must be a positive number)' read(*,*) a if(a>=0.0) then roota = sqrt(a) write(*,*) 'The sqaure root of ', a, ' = ', roota pause end if stop end program emosix
 P: 3 Hi Thanks for the response, its in there to stop the program if the user puts a negative number in, as you can't have a square root of negative number. I put in the end if and it gives me another error if(a>=0.0) 1 Error: Cannot assign to a named constant at (1) C:\Users\James\Documents\Programming\Emo6\~emo6.f90:11.3: end if 1 Error: Expecting END PROGRAM statement at (1) C:\Users\James\Documents\Programming\Emo6\~emo6.f90:17.5: pause 1 Warning: Deleted feature: PAUSE statement at (1) that is what I get now.
P: 3

## Won't Compile the simplest of programs! Am I being stupid???

oops its supposed to be 'a<=0.0' but that didn't change anything.

Thanks

Jim
Mentor
P: 19,859
 Quote by JamesWMH Hi Thanks for the response, its in there to stop the program if the user puts a negative number in, as you can't have a square root of negative number.
Well, you can, but the result won't be real. Instead of stopping the program, you should have logic that prevents the program from attempting to take the square root of a negative number.
 Quote by JamesWMH I put in the end if and it gives me another error if(a>=0.0) 1 Error: Cannot assign to a named constant at (1) C:\Users\James\Documents\Programming\Emo6\~emo6.f90:11.3: end if 1 Error: Expecting END PROGRAM statement at (1) C:\Users\James\Documents\Programming\Emo6\~emo6.f90:17.5: pause 1 Warning: Deleted feature: PAUSE statement at (1) that is what I get now.
 Quote by JamesWMH oops its supposed to be 'a<=0.0' but that didn't change anything.
You can take the square root of 0, so you should make the test expression a < 0.0.

As far as the "can't assign to a named constant" business, try using a different variable for roota.

What does your code look like now? You if statement wasn't formed correctly before, and it still might not be right.
Mentor
P: 10,729
 Quote by JamesWMH if(a>=0.0)
You need a "then". For example (also fixing your test condition):

      if (a <= 0.0) then
stop
end if
For some reason, the missing "then" triggers the "can't assign to a named constant" error.
Mentor
P: 19,859
 Quote by jtbell You need a "then".
I added one in the code I showed in post #2, but didn't say that explicitly. I should have, since the OP seems to have missed it.
 Quote by jtbell For example (also fixing your test condition):  if (a <= 0.0) then stop end if For some reason, the missing "then" triggers the "can't assign to a named constant" error.
The condition can be written as a < 0.0.
 P: 2 suggestion of mark is absolutely correct the error is coming due to absence of THEN statement thanx
 Sci Advisor P: 2,728 I know this is an old thread that's been bumped for some reason, but I would like to add one more thing. The Fortran "if" statement doesn't require a "then" (or an "endif" either) when the "if" is only controlling a single statement. In this sense the OP's code should (almost) have worked correctly, had he not put the "new line" after the "if" statement (since end of line is more than just white space in Fortran source code).  if (a <= 0.0) stop This should have worked correctly without "then" and without "endif"

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