我试图从生锈中的C函数中检索原始指针，并使用相同的原始指针作为另一个库中另一个C函数的参数。当我传递原始指针时，我最终在C端有一个NULL指针。
我试图制作一个简化版本的问题，但是当我这样做的时候就像我期望的那样 -
C代码 -
struct MyStruct {
int value;
};
struct MyStruct * get_struct() {
struct MyStruct * priv_struct = (struct MyStruct*) malloc( sizeof(struct MyStruct));
priv_struct->value = 0;
return priv_struct;
}
void put_struct(struct MyStruct *priv_struct) {
printf("Value - %d\n", priv_struct->value);
}
Rust Code -
#[repr(C)]
struct MyStruct {
value: c_int,
}
extern {
fn get_struct() -> *mut MyStruct;
}
extern {
fn put_struct(priv_struct: *mut MyStruct) -> ();
}
fn rust_get_struct() -> *mut MyStruct {
let ret = unsafe { get_struct() };
ret
}
fn rust_put_struct(priv_struct: *mut MyStruct) {
unsafe { put_struct(priv_struct) };
}
fn main() {
let main_struct = rust_get_struct();
rust_put_struct(main_struct);
}
当我运行它时，我得到Value - 0
的输出~/Dev/rust_test$ sudo ./target/debug/rust_test
Value - 0
~/Dev/rust_test$
但是，当尝试对DPDK库执行此操作时，我以相同的方式检索并传递原始指针，但得到段错误。如果我使用gdb进行调试，我可以看到我在Rust端传递指针，但我在C端看到它为NULL -
(gdb) frame 0
#0 rte_eth_rx_queue_setup (port_id=0 '\000', rx_queue_id=<optimized out>, nb_rx_desc=<optimized out>, socket_id=0, rx_conf=0x0, mp=0x0)
at /home/kenton/Dev/dpdk-16.07/lib/librte_ether/rte_ethdev.c:1216
1216 if (mp->private_data_size < sizeof(struct rte_pktmbuf_pool_private)) {
(gdb) frame 1
#1 0x000055555568953b in dpdk::ethdev::dpdk_rte_eth_rx_queue_setup (port_id=0 '\000', rx_queue_id=0, nb_tx_desc=128, socket_id=0, rx_conf=None,
mb=0x7fff3fe47640) at /home/kenton/Dev/dpdk_ffi/src/ethdev/mod.rs:32
32 let retc: c_int = unsafe {ffi::rte_eth_rx_queue_setup(port_id as uint8_t,
在第1帧中， mb 有一个地址并正在传递。在第0帧中，库中的接收函数将 mp 显示为0x0。
我接收指针的代码 -
let mb = dpdk_rte_pktmbuf_pool_create(CString::new("MBUF_POOL").unwrap().as_ptr(),
(8191 * nb_ports) as u32 , 250, 0, 2176, dpdk_rte_socket_id());
调用ffi库 -
pub fn dpdk_rte_pktmbuf_pool_create(name: *const c_char,
n: u32,
cache_size: u32,
priv_size: u16,
data_room_size: u16,
socket_id: i32) -> *mut rte_mempool::ffi::RteMempool {
let ret: *mut rte_mempool::ffi::RteMempool = unsafe {
ffi::shim_rte_pktmbuf_pool_create(name,
n as c_uint,
cache_size as c_uint,
priv_size as uint16_t,
data_room_size as uint16_t,
socket_id as c_int)
};
ret
}
ffi -
extern {
pub fn shim_rte_pktmbuf_pool_create(name: *const c_char,
n: c_uint,
cache_size: c_uint,
priv_size: uint16_t,
data_room_size: uint16_t,
socket_id: c_int) -> *mut rte_mempool::ffi::RteMempool;
}
C函数 -
struct rte_mempool *
rte_pktmbuf_pool_create(const char *name, unsigned n,
unsigned cache_size, uint16_t priv_size, uint16_t data_room_size,
int socket_id);
当我传递指针时，它看起来与上面的简化版本大致相同。我的变量 mb 包含一个原始指针，我将其传递给另一个函数 -
ret = dpdk_rte_eth_rx_queue_setup(port,q,128,0,None,mb);
ffi library -
pub fn dpdk_rte_eth_rx_queue_setup(port_id: u8,
rx_queue_id: u16,
nb_tx_desc: u16,
socket_id: u32,
rx_conf: Option<*const ffi::RteEthRxConf>,
mb_pool: *mut rte_mempool::ffi::RteMempool ) -> i32 {
let retc: c_int = unsafe {ffi::rte_eth_rx_queue_setup(port_id as uint8_t,
rx_queue_id as uint16_t,
nb_tx_desc as uint16_t,
socket_id as c_uint,
rx_conf,
mb)};
let ret: i32 = retc as i32;
ret
}
ffi -
extern {
pub fn rte_eth_rx_queue_setup(port_id: uint8_t,
rx_queue_id: uint16_t,
nb_tx_desc: uint16_t,
socket_id: c_uint,
rx_conf: Option<*const RteEthRxConf>,
mb: *mut rte_mempool::ffi::RteMempool ) -> c_int;
}
C函数 -
int
rte_eth_rx_queue_setup(uint8_t port_id, uint16_t rx_queue_id,
uint16_t nb_rx_desc, unsigned int socket_id,
const struct rte_eth_rxconf *rx_conf,
struct rte_mempool *mp);
我为这个长度道歉，但我觉得我错过了一些简单的东西，并且我们无法弄明白。我已经检查了正在传递的每个字段的结构对齐，我甚至看到了接收到的指针的值，因为我期待 -
(gdb) frame 1
#1 0x000055555568dcf4 in dpdk::ethdev::dpdk_rte_eth_rx_queue_setup (port_id=0 '\000', rx_queue_id=0, nb_tx_desc=128, socket_id=0, rx_conf=None,
mb=0x7fff3fe47640) at /home/kenton/Dev/dpdk_ffi/src/ethdev/mod.rs:32
32 let retc: c_int = unsafe {ffi::rte_eth_rx_queue_setup(port_id as uint8_t,
(gdb) print *mb
$1 = RteMempool = {name = "MBUF_POOL", '\000' <repeats 22 times>, pool_union = PoolUnionStruct = {data = 140734245862912}, pool_config = 0x0,
mz = 0x7ffff7fa4c68, flags = 16, socket_id = 0, size = 8191, cache_size = 250, elt_size = 2304, header_size = 64, trailer_size = 0,
private_data_size = 64, ops_index = 0, local_cache = 0x7fff3fe47700, populated_size = 8191, elt_list = RteMempoolObjhdrList = {
stqh_first = 0x7fff3ebc7f68, stqh_last = 0x7fff3fe46ce8}, nb_mem_chunks = 1, mem_list = RteMempoolMemhdrList = {stqh_first = 0x7fff3ebb7d80,
stqh_last = 0x7fff3ebb7d80}, __align = 0x7fff3fe47700}
关于为什么指针在C端变为NULL的任何想法？
答案 0 :(得分：3)
CString::new("…").unwrap().as_ptr()不起作用。 CString是临时的，因此as_ptr()调用返回该临时的内部指针，当您使用它时可能会悬空。只要你不使用指针，每个Rust的安全定义都是“安全的”，但你最终会在unsafe块中这样做。您应该将字符串绑定到变量，并对该变量使用as_ptr。
这是一个常见的问题，甚至有a proposal to fix the CStr{,ing} API to avoid it。
此外，原始指针本身可以为空，因此相当于const struct rte_eth_rxconf *的Rust FFI将为*const ffi::RteEthRxConf，而不是Option<*const ffi::RteEthRxConf>。