Error cannot assign to a named constant at 1

if (z.le.zstart) then
   if (y.ge.((6.95*wg_y2)/5)).and.(y.le.((12.55*wg_y2)/5)) then
      indexmedia=nd
   end if
end if

0

Description

Sergio Losilla

2014-02-11 13:29:09 UTC

Compiler version:
GNU Fortran (Ubuntu 20130917-1ubuntu1) 4.9.0 20130917 (experimental) [trunk revision 202647]

Test program:
--------------------
program ifelif
    if (.TRUE.)
    else if (.FALSE.)
    end if
end program
--------------------

Error messages:
----------------------
if.F90:2.15:

    if (.TRUE.)
               1
Error: Cannot assign to a named constant at (1)
if.F90:3.11:

    else if (.FALSE.)
           1
Error: Unexpected junk after ELSE statement at (1)
[...]
----------------------

Both errors should read something like "Missing THEN after IF/ELSE IF".

The first error message doesn't even make sense...

Comment 1

kargl

2014-02-11 14:35:07 UTC

(In reply to Sergio Losilla from comment #0)
> Compiler version:
> GNU Fortran (Ubuntu 20130917-1ubuntu1) 4.9.0 20130917 (experimental) [trunk
> revision 202647]
> 
> Test program:
> --------------------
> program ifelif
>     if (.TRUE.)
>     else if (.FALSE.)
>     end if
> end program
> --------------------
> 
> Error messages:
> ----------------------
> if.F90:2.15:
> 
>     if (.TRUE.)
>                1
> Error: Cannot assign to a named constant at (1)
> if.F90:3.11:
> 
>     else if (.FALSE.)
>            1
> Error: Unexpected junk after ELSE statement at (1)
> [...]
> ----------------------
> 
> Both errors should read something like "Missing THEN after IF/ELSE IF".
> 
> The first error message doesn't even make sense...


How is the compiler suppose to know that the programmer may have
meant

   program ifelif
   if (.TRUE.) i = 42
   if (.FALSE.) then
      j = 42
   end if
   end program

As for the message that you think makes no sense, I suspect
the parse has identified (.TRUE.) as a malformed expression.

Comment 2

Sergio Losilla

2014-02-11 14:50:18 UTC

(In reply to kargl from comment #1)
> How is the compiler suppose to know that the programmer may have
> meant
> 
>    program ifelif
>    if (.TRUE.) i = 42
>    if (.FALSE.) then
>       j = 42
>    end if
>    end program

As far as I know, there are two possible statements starting with if:

if (<expr>) then
if (<expr>) <statement>

The compiler should recognize that *something* is missing, be it either a then or a statement.

> As for the message that you think makes no sense, I suspect
> the parse has identified (.TRUE.) as a malformed expression.

Well, the message objectively makes no sense because there is no assignment whatsoever. At the very least, the message is confusing.

Comment 3

Dominique d’Humieres

2014-02-11 18:30:59 UTC

Confirmed on 4.7.4, 4.8.3, and trunk (4.9). Related to pr32151.

Comment 4

Dominique d’Humieres

2016-02-25 13:57:56 UTC

*** Bug 38303 has been marked as a duplicate of this bug. ***

Comment 5

Dominique d’Humieres

2019-02-02 19:32:42 UTC

> Error messages:
> ----------------------
> if.F90:2.15:
>
>     if (.TRUE.)
>                1
> Error: Cannot assign to a named constant at (1)
> if.F90:3.11:
>
>     else if (.FALSE.)
>            1
> Error: Unexpected junk after ELSE statement at (1)
> [...]
> ----------------------
>
> Both errors should read something like "Missing THEN after IF/ELSE IF".
>
> The first error message doesn't even make sense...

% gfc dec_type_print_red.f90
dec_type_print_red.f90:20:11:

   20 |     if (b) type*,a(i) ! test TYPE as PRINT inside one-line IF
      |           1
Error: Cannot assign to a named constant at (1)

(from gfortran.dg/dec_type_print.f90 compiled without the -fdec option).

IMO replacing "Cannot assign to a named constant" with "Syntax error" will give an immediate clue of what is wrong. A longer alternative could be something such as
""Syntax error, missing or bad STATEMENT".

Comment 6

Dominique d’Humieres

2019-02-19 13:10:55 UTC

I don't understand the goal of the block (line 1651 of gcc/fortran/match.c)

  /* The gfc_match_assignment() above may have returned a MATCH_NO
     where the assignment was to a named constant.  Check that
     special case here.  */
  m = gfc_match_assignment ();
  if (m == MATCH_NO)
   {
      gfc_error ("Cannot assign to a named constant at %C");
      gfc_free_expr (expr);
      gfc_undo_symbols ();
      gfc_current_locus = old_loc;
      return MATCH_ERROR;
   }

The closest test matching my understanding of the comment is

integer, parameter :: i = 1
if (i == 2) i = 3 ! { dg-error "in variable definition context" }
end

which gives the following error

if_param.f90:2:12:

    2 | if (i == 2) i = 3 ! { dg-error "in variable definition context" }
      |            1
Error: Named constant 'i' in variable definition context (assignment) at (1)

If I remove the block, I get the following errors for the test in comment 0:

pr60144.f90:2:15:

    2 |     if (.TRUE.)
      |               1
Error: Unclassifiable statement in IF-clause at (1)
pr60144.f90:3:11:

    3 |     else if (.FALSE.)
      |           1
Error: Unexpected junk after ELSE statement at (1)

which looks better for the IF statement.

For the ELSE IF error, how can I move the caret after '(.FALSE.)' and replace 'junk' with something neutral (token, statement, ...)?

Comment 7

dominiq

2019-05-01 17:40:58 UTC

Author: dominiq
Date: Wed May  1 17:40:22 2019
New Revision: 270776

URL: https://gcc.gnu.org/viewcvs?rev=270776&root=gcc&view=rev
Log:
2019-05-01 Dominique d'Humieres  <dominiq@gcc.gnu.org>

	PR fortran/60144
	* match.c (gfc_match_parens): Change the location for missing ')'.
	(gfc_match_if): Detect a missing '('. Remove the spurious named
	constant error. Change the wording of some errors.
	(gfc_match_else): Change the wording of an error.
	(gfc_match_elseif): Detect a missing '('. Improve the matching
	process to get a better syntax analysis.

	PR fortran/60144
	* gfortran.dg/block_name_2.f90: Adjust dg-error.
	* gfortran.dg/dec_type_print_3.f90.f90: Likewise
	* gfortran.dg/pr60144.f90: New test. 


Added:
    trunk/gcc/testsuite/gfortran.dg/pr60144.f90
Modified:
    trunk/gcc/fortran/ChangeLog
    trunk/gcc/fortran/match.c
    trunk/gcc/testsuite/ChangeLog
    trunk/gcc/testsuite/gfortran.dg/block_name_2.f90
    trunk/gcc/testsuite/gfortran.dg/dec_type_print_3.f90

Comment 8

Dominique d’Humieres

2020-07-30 15:01:33 UTC

This is fixed since at least GCC7.

Александр Звягинцев

Добрый день.
Почему такое решение дает ошибку:

// BEGIN (write your solution here)
const finalGrade = (exam, projects) => {
if (exam > 90 || projects > 10) {
finalGrade= 100;
} else if (exam > 75 && projects >= 5) {
finalGrade= 90;
} else if (exam > 50 && projects >= 2) {
finalGrade= 75;
} else {
finalGrade= 0;
}
return finalGrade;
}
// END

А такое нет:

// BEGIN
const finalGrade = (exam, projects) => {
if (exam > 90 || projects > 10) {
return 100;
} else if (exam > 75 && projects >= 5) {
return 90;
} else if (exam > 50 && projects >= 2) {
return 75;
} else {
return 0;
}
};
// END

Я думал по сути это одно и тоже.

0

TypeError: Assignment to constant variable.

  at finalGrade (finalGrade.js:4:16)
  at Object.<anonymous>.test (__tests__/finalGrade.test.js:4:35)
  at Promise.resolve.then.el (../../local/share/.config/yarn/global/node_modules/p-map/index.js:42:16)

0

0

0

0

У меня сложилось впечатление, что определения макросов в стиле C работают в gfortran?

#define ERROR_CHECKER(msg,stat) 
      IF (stat.NE.0) THEN 
         DO i = 1,BIG 
            IF(msg(i).NE.C_NULL_CHAR)THEN 
               ErrMsg(i:i) = msg(i) 
            ELSE 
               EXIT 
            END IF 
         END DO
         IF(stat.EQ.1) THEN 
            ErrStat = ID_Warn 
         ELSE 
            ErrStat = ID_Fatal 
            RETURN 
        END IF 
     END IF

Но затем эта ошибка портит мне день:

IF (stat.NE.0) THEN      DO i = 1,BIG         IF(message
               1
Error: Cannot assign to a named constant at (1)

Есть идеи, что я здесь делаю неправильно?

Второй вопрос: распознает ли Intel fortran макросы c-style? Если да, нужны ли флаги компилятора?

Вместо использования макроса просто преобразуйте макрос в функцию. Таким образом, у вас не будет большой зависимости от компилятора Fortran, имеющего возможность макроса.

LOGICAL FUNCTION ERROR_CHECKER(msg,stat)
  character*(*) msg(*)
  integer stat
  IF (stat.NE.0) THEN
     DO i = 1,BIG
        IF(msg(i).NE.C_NULL_CHAR)THEN
           ErrMsg(i:i) = msg(i)
        ELSE
           EXIT
        END IF
     END DO
     IF(stat.EQ.1) THEN
        ErrStat = ID_Warn
     ELSE
        ErrStat = ID_Fatal
        RETURN .FALSE.
    END IF
 END IF
 RETURN .TRUE.
 END FUNCTION

В вашем коде

IF (ERROR_CHECKER(msg, stat)) RETURN

Изменить: в некоторых более поздних версиях Fortran есть разделители операторов (;), которые можно использовать. К сожалению, длина строки ограничена, поэтому ваш макрос не может быть очень длинным и содержать более одной управляющей структуры.

Помимо того факта, что такой подход к макросам не является хорошим стилем, вам не хватает необходимых разрывов строк в сгенерированном коде. Это можно исправить, поставив точку с запятой перед каждой обратной косой чертой в определении макроса (кроме первой строки).