Refactor: simplify an if statement #4281
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thk123
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kroening,
martin-cs,
peterschrammel and
tautschnig
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|
Note if you don't like option 2, note the first commit is less ambitious in this trivial tidy up. |
| return false; | ||
| const bool is_dynamic = expr.type().get_bool(ID_C_dynamic); | ||
| const auto symbol_expr = expr_try_dynamic_cast<symbol_exprt>(expr); | ||
| return is_dynamic && symbol_expr && |
tautschnig
approved these changes
Feb 26, 2019
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|
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| return expr.id() == ID_symbol && | ||
| const bool is_dynamic = expr.type().get_bool(ID_C_dynamic); | ||
| const auto symbol_expr = expr_try_dynamic_cast<symbol_exprt>(expr); |
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Apart from the bug that @romainbrenguier spotted, this approach has the downside that we do extra work that may never be necessary in case is_dynamic is true. Could we just do
return expr.type().get_bool(ID_C_dynamic) || (... return statement from the first commit ...);
?
I don't think this is super performance critical, but I don't think this is a massive improvement either.
Without branching it is simpler to see what the condition is
thk123
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Done as requested. |
allredj
reviewed
Feb 27, 2019
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✔️
Passed Diffblue compatibility checks (cbmc commit: 86b5e11).
Build URL: https://travis-ci.com/diffblue/test-gen/builds/102469149
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Spotted this doing #3481 - if anyone doesn't like this, I'll simply drop it.