In this example: nft --debug=netlink add rule ip t c ip dscp set ip dscp ip t c [ payload load 2b @ network header + 0 => reg 1 ] [ bitwise reg 1 = (reg=1 & 0x000003ff ) ^ 0x00000000 ] [ payload load 1b @ network header + 1 => reg 2 ] [ bitwise reg 2 = (reg=2 & 0x0000003c ) ^ 0x00000000 ] [ bitwise reg 2 = ( reg 2 >> 0x00000002 ) ] [ bitwise reg 2 = ( reg 2 << 0x00000002 ) ] [ bitwise reg 1 = (reg=1 & 0x0000ffff ) ^ reg 2 ] [ payload write reg 1 => 2b @ network header + 0 csum_type 1 csum_off 10 csum_flags 0x0 ] The mask at line 4 should be 0xfc, not 0x3c. Evaluation of the payload expression munges it from `ip dscp` to `(ip dscp & 0xfc) >> 2`, because although `ip dscp` is only 6 bits long, those 6 bits are the top bits in a byte, and to make the arithmetic simpler when we perform comparisons and assignments, we mask and shift the field. When the AND expression is allocated, its length is correctly set to 8. However, when a binop is evaluated, it is assumed that the length has not been set and is length is always set to the length of the left operand, incorrectly to 6 in this case. When the bitwise netlink expression is generated, the length of the AND is used to generate the mask, 0x3f, used in combining the binop's. The upshot of this is that the original mask gets mangled to 0x3c. We can fix this by changing the evaluation of binops only to set the op's length if it is not already set. Signed-off-by: Jeremy Sowden <jeremy@xxxxxxxxxx> --- src/evaluate.c | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/src/evaluate.c b/src/evaluate.c index 1db175007c2d..ff168434cd8f 100644 --- a/src/evaluate.c +++ b/src/evaluate.c @@ -1133,7 +1133,7 @@ static int expr_evaluate_bitwise(struct eval_ctx *ctx, struct expr **expr) op->dtype = left->dtype; op->byteorder = left->byteorder; - op->len = left->len; + op->len = op->len ? op->len : left->len; if (expr_is_constant(left) && expr_is_constant(op->right)) return constant_binop_simplify(ctx, expr); -- 2.25.1
