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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> |
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lxdialog | ||
tests | ||
.gitignore | ||
array_size.h | ||
conf.c | ||
confdata.c | ||
expr.c | ||
expr.h | ||
gconf-cfg.sh | ||
gconf.c | ||
gconf.glade | ||
hashtable.h | ||
images.c | ||
images.h | ||
internal.h | ||
lexer.l | ||
list_types.h | ||
list.h | ||
lkc_proto.h | ||
lkc.h | ||
Makefile | ||
mconf-cfg.sh | ||
mconf.c | ||
menu.c | ||
merge_config.sh | ||
mnconf-common.c | ||
mnconf-common.h | ||
nconf-cfg.sh | ||
nconf.c | ||
nconf.gui.c | ||
nconf.h | ||
parser.y | ||
preprocess.c | ||
preprocess.h | ||
qconf-cfg.sh | ||
qconf.cc | ||
qconf.h | ||
streamline_config.pl | ||
symbol.c | ||
util.c |