[ Upstream commit 77a92660d8 ]
expr_trans_bool() performs an incorrect transformation.
[Test Code]
config MODULES
def_bool y
modules
config A
def_bool y
select C if B != n
config B
def_tristate m
config C
tristate
[Result]
CONFIG_MODULES=y
CONFIG_A=y
CONFIG_B=m
CONFIG_C=m
This output is incorrect because CONFIG_C=y is expected.
Documentation/kbuild/kconfig-language.rst clearly explains the function
of the '!=' operator:
If the values of both symbols are equal, it returns 'n',
otherwise 'y'.
Therefore, the statement:
select C if B != n
should be equivalent to:
select C if y
Or, more simply:
select C
Hence, the symbol C should be selected by the value of A, which is 'y'.
However, expr_trans_bool() wrongly transforms it to:
select C if B
Therefore, the symbol C is selected by (A && B), which is 'm'.
The comment block of expr_trans_bool() correctly explains its intention:
* bool FOO!=n => FOO
^^^^
If FOO is bool, FOO!=n can be simplified into FOO. This is correct.
However, the actual code performs this transformation when FOO is
tristate:
if (e->left.sym->type == S_TRISTATE) {
^^^^^^^^^^
While it can be fixed to S_BOOLEAN, there is no point in doing so
because expr_tranform() already transforms FOO!=n to FOO when FOO is
bool. (see the "case E_UNEQUAL" part)
expr_trans_bool() is wrong and unnecessary.
Signed-off-by: Masahiro Yamada <masahiroy@kernel.org>
Acked-by: Randy Dunlap <rdunlap@infradead.org>
Signed-off-by: Sasha Levin <sashal@kernel.org>
da0fec30a7