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>