Dear all,

in find_substring_ref we erroneously handled given substring start and end
indices as unsigned integers.  However, gives indices could be negative,
which is legal as long as end < start, leading to a string of length zero.
The current behavior could lead to a wrong length as well as an invalid read
from (compiler) memory.

The fix allows to reintroduce code in testcase substr_6.f90 that was
erroneously considered as illegal.

Regtested on x86_64-pc-linux-gnu.  OK for mainline?

As this is invalid code, I'd like to backport this fix.

Thanks,
Harald


Fortran - fix handling of substring start and end indices

gcc/fortran/ChangeLog:

        PR fortran/85130
        * expr.c (find_substring_ref): Handle given substring start and
        end indices as signed integers, not unsigned.

gcc/testsuite/ChangeLog:

        PR fortran/85130
        * gfortran.dg/substr_6.f90: Revert commit r8-7574, adding again
        test that was erroneously considered as illegal.

diff --git a/gcc/fortran/expr.c b/gcc/fortran/expr.c
index dfecc3012e1..604e63e6164 100644
--- a/gcc/fortran/expr.c
+++ b/gcc/fortran/expr.c
@@ -1724,8 +1724,8 @@ find_substring_ref (gfc_expr *p, gfc_expr **newp)
   *newp = gfc_copy_expr (p);
   free ((*newp)->value.character.string);

-  end = (gfc_charlen_t) mpz_get_ui (p->ref->u.ss.end->value.integer);
-  start = (gfc_charlen_t) mpz_get_ui (p->ref->u.ss.start->value.integer);
+  end = (gfc_charlen_t) mpz_get_si (p->ref->u.ss.end->value.integer);
+  start = (gfc_charlen_t) mpz_get_si (p->ref->u.ss.start->value.integer);
   if (end >= start)
     length = end - start + 1;
   else
diff --git a/gcc/testsuite/gfortran.dg/substr_6.f90 b/gcc/testsuite/gfortran.dg/substr_6.f90
index 0d5e3d75e88..83e788a55a6 100644
--- a/gcc/testsuite/gfortran.dg/substr_6.f90
+++ b/gcc/testsuite/gfortran.dg/substr_6.f90
@@ -6,6 +6,8 @@ CHARACTER(5), parameter :: c0(1) = (/ "123" // ACHAR(0) // "5" /)
 CHARACTER*5 c(1)
 CHARACTER(1), parameter :: c1(5) = (/ "1", "2", "3", ACHAR(0), "5" /)

+c = c0(1)(-5:-8)
+if (c(1) /= "     ") STOP 1
 c = (/ c0(1)(1:5) /)
 do i=1,5
    if (c(1)(i:i) /= c1(i)) STOP 2

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