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59bf49a632
When a statement function in a nested scope has a name that clashes with a name that exists in the host scope, the compiler can handle it correctly (with a portability warning)... unless the host scope acquired the name via USE association. Fix. Fixes https://github.com/llvm/llvm-project/issues/88678.
59 lines
1.7 KiB
Fortran
59 lines
1.7 KiB
Fortran
! RUN: %python %S/test_errors.py %s %flang_fc1 -pedantic
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module m1
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contains
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real function rf2(x)
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rf2 = x
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end
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end
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module m2
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use m1
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real, target :: x = 1.
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contains
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function rpf(x)
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real, intent(in out), target :: x
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real, pointer :: rpf
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rpf => x
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end
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real function rf(x)
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rf = x
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end
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subroutine test1
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! This is a valid assignment, not a statement function.
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! Every other Fortran compiler misinterprets it!
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rpf(x) = 2. ! statement function or indirect assignment?
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print *, x
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end
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subroutine test2
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!PORTABILITY: Name 'rf' from host scope should have a type declaration before its local statement function definition
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rf(x) = 1.
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end
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subroutine test2b
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!PORTABILITY: Name 'rf2' from host scope should have a type declaration before its local statement function definition
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rf2(x) = 1.
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end
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subroutine test3
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external sf
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!ERROR: 'sf' has not been declared as an array or pointer-valued function
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sf(x) = 4.
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end
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function f()
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!ERROR: Recursive call to 'f' requires a distinct RESULT in its declaration
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!ERROR: Left-hand side of assignment is not definable
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!BECAUSE: 'f()' is not a variable or pointer
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f() = 1. ! statement function of same name as function
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end
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function g() result(r)
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!WARNING: Name 'g' from host scope should have a type declaration before its local statement function definition
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!ERROR: 'g' is already declared in this scoping unit
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g() = 1. ! statement function of same name as function
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end
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function h1() result(r)
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!ERROR: 'r' is not a callable procedure
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r() = 1. ! statement function of same name as function result
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end
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function h2() result(r)
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procedure(real), pointer :: r
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r() = 1. ! not a statement function
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end
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end
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