File: i40e_xsk.h

package info (click to toggle)
linux 6.1.4-1
  • links: PTS, VCS
  • area: main
  • in suites: bookworm
  • size: 1,532,052 kB
  • sloc: ansic: 23,400,063; asm: 266,720; sh: 108,896; makefile: 49,712; python: 36,925; perl: 36,810; cpp: 6,044; yacc: 4,904; lex: 2,722; awk: 1,440; ruby: 25; sed: 5
file content (38 lines) | stat: -rw-r--r-- 1,360 bytes parent folder | download | duplicates (9)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
/* SPDX-License-Identifier: GPL-2.0 */
/* Copyright(c) 2018 Intel Corporation. */

#ifndef _I40E_XSK_H_
#define _I40E_XSK_H_

/* This value should match the pragma in the loop_unrolled_for
 * macro. Why 4? It is strictly empirical. It seems to be a good
 * compromise between the advantage of having simultaneous outstanding
 * reads to the DMA array that can hide each others latency and the
 * disadvantage of having a larger code path.
 */
#define PKTS_PER_BATCH 4

#ifdef __clang__
#define loop_unrolled_for _Pragma("clang loop unroll_count(4)") for
#elif __GNUC__ >= 8
#define loop_unrolled_for _Pragma("GCC unroll 4") for
#else
#define loop_unrolled_for for
#endif

struct i40e_vsi;
struct xsk_buff_pool;

int i40e_queue_pair_disable(struct i40e_vsi *vsi, int queue_pair);
int i40e_queue_pair_enable(struct i40e_vsi *vsi, int queue_pair);
int i40e_xsk_pool_setup(struct i40e_vsi *vsi, struct xsk_buff_pool *pool,
			u16 qid);
bool i40e_alloc_rx_buffers_zc(struct i40e_ring *rx_ring, u16 cleaned_count);
int i40e_clean_rx_irq_zc(struct i40e_ring *rx_ring, int budget);

bool i40e_clean_xdp_tx_irq(struct i40e_vsi *vsi, struct i40e_ring *tx_ring);
int i40e_xsk_wakeup(struct net_device *dev, u32 queue_id, u32 flags);
int i40e_realloc_rx_bi_zc(struct i40e_vsi *vsi, bool zc);
void i40e_clear_rx_bi_zc(struct i40e_ring *rx_ring);

#endif /* _I40E_XSK_H_ */