Andrzej Warzynski 96d229c9ab [flang][driver] Add options for unparsing
This patch adds the following compiler frontend driver options:
  * -fdebug-unparse (f18 spelling: -funparse)
  * -fdebug-unparse-with-symbols (f18 spelling: -funparse-with-symbols)
The new driver will only accept the new spelling. `f18` will accept both
the original and the new spelling.

A new base class for frontend actions is added: `PrescanAndSemaAction`.
This is added to reduce code duplication that otherwise these new
options would lead to. Implementation from
  * `ParseSyntaxOnlyAction::ExecutionAction`
is moved to:
  * `PrescanAndSemaAction::BeginSourceFileAction`
This implementation is now shared between:
  * PrescanAndSemaAction
  * ParseSyntaxOnlyAction
  * DebugUnparseAction
  * DebugUnparseWithSymbolsAction

All tests that don't require other yet unimplemented options are
updated. This way `flang-new -fc1` is used instead of `f18` when
`FLANG_BUILD_NEW_DRIVER` is set to `On`. In order to facilitate this,
`%flang_fc1` is added in the LIT configuration (lit.cfg.py).

`asFortran` from f18.cpp is duplicated as `getBasicAsFortran` in
FrontendOptions.cpp. At this stage it's hard to find a good place to
share this method. I suggest that we revisit this once a switch from
`f18` to `flang-new` is complete.

Differential Revision: https://reviews.llvm.org/D96483
2021-02-16 09:32:51 +00:00

56 lines
1.3 KiB
Fortran

! RUN: %flang_fc1 -fdebug-unparse-with-symbols %s 2>&1 | FileCheck %s
! CHECK-NOT: do [1-9]
! Figure out how to also execute this test.
program main
integer :: results(100)
integer :: count
count = 0
if (.true.) then
do 1 j1=1,2
count = count + 1
results(count) = j1
1 continue
end if
do 2 j1=3,4
do 2 j2=1,2
if (j1 == j2) then
do 3 j3=1,2
count = count + 1
results(count) = 100*j1 + 10*j2 + j3
do 3 j4=1,2
do
count = count + 1
results(count) = 10*j3 + j4
exit
end do
3 end do
else
do
do 4 j3=3,4
count = count + 1
results(count) = 100*j1 + 10*j2 + j3
do 4 j4=3,4
count = count + 1
results(count) = 10*j3 + j4
4 end do
exit
end do
end if
count = count + 1
results(count) = 10*j1 + j2
2 continue
do 5 j1=5,6 ! adjacent non-block DO loops
count = count + 1
5 results(count) = j1
do 6 j1=7,8 ! non-block DO loop at end of execution part
count = count + 1
6 results(count) = j1
if (count == 34 .and. sum(results(1:count)) == 3739) then
print *, 'pass'
else
print *, 'FAIL:', count, sum(results(1:count)), results(1:count)
end if
end