Hi! This optimization was written purely with scalar integers in mind, can work fine even with vectors, but we can't use build_int_cst but need to use build_one_cst instead.
Bootstrapped/regtested on x86_64-linux and i686-linux, ok for trunk? 2021-02-24 Jakub Jelinek <ja...@redhat.com> PR tree-optimization/99225 * fold-const.c (fold_binary_loc) <case NE_EXPR>: In (x & (1 << y)) != 0 to ((x >> y) & 1) != 0 simplifications use build_one_cst instead of build_int_cst (..., 1). Formatting fixes. * gcc.c-torture/compile/pr99225.c: New test. --- gcc/fold-const.c.jj 2021-02-23 09:49:40.000000000 +0100 +++ gcc/fold-const.c 2021-02-23 19:53:33.143763292 +0100 @@ -12044,23 +12044,23 @@ fold_binary_loc (location_t loc, enum tr && integer_onep (TREE_OPERAND (arg00, 0))) { tree tem = fold_build2_loc (loc, RSHIFT_EXPR, TREE_TYPE (arg00), - arg01, TREE_OPERAND (arg00, 1)); + arg01, TREE_OPERAND (arg00, 1)); tem = fold_build2_loc (loc, BIT_AND_EXPR, TREE_TYPE (arg0), tem, - build_int_cst (TREE_TYPE (arg0), 1)); + build_one_cst (TREE_TYPE (arg0))); return fold_build2_loc (loc, code, type, - fold_convert_loc (loc, TREE_TYPE (arg1), tem), - arg1); + fold_convert_loc (loc, TREE_TYPE (arg1), + tem), arg1); } else if (TREE_CODE (arg01) == LSHIFT_EXPR && integer_onep (TREE_OPERAND (arg01, 0))) { tree tem = fold_build2_loc (loc, RSHIFT_EXPR, TREE_TYPE (arg01), - arg00, TREE_OPERAND (arg01, 1)); + arg00, TREE_OPERAND (arg01, 1)); tem = fold_build2_loc (loc, BIT_AND_EXPR, TREE_TYPE (arg0), tem, - build_int_cst (TREE_TYPE (arg0), 1)); + build_one_cst (TREE_TYPE (arg0))); return fold_build2_loc (loc, code, type, - fold_convert_loc (loc, TREE_TYPE (arg1), tem), - arg1); + fold_convert_loc (loc, TREE_TYPE (arg1), + tem), arg1); } } --- gcc/testsuite/gcc.c-torture/compile/pr99225.c.jj 2021-02-23 20:12:01.825464969 +0100 +++ gcc/testsuite/gcc.c-torture/compile/pr99225.c 2021-02-23 20:11:45.962640464 +0100 @@ -0,0 +1,31 @@ +/* PR tree-optimization/99225 */ + +typedef int V __attribute__((vector_size (4 * sizeof (int)))); + +void +foo (V *x) +{ + x[2] = (x[0] & (1 << x[1])) != 0; +} + +void +bar (V *x) +{ + x[2] = ((1 << x[1]) & x[0]) != 0; +} + +void +baz (V *x) +{ + V a = 1 << x[1]; + V b = a & x[0]; + x[2] = b != 0; +} + +void +qux (V *x) +{ + V a = 1 << x[1]; + V b = x[0] & a; + x[2] = b != 0; +} Jakub