我正在使用 netif 结构(类似于 http://www.nongnu.org/lwip/structnetif.html),我遇到了与对齐相关的问题。我注意到每个 int 都以 4 的乘数地址开始(例如 0x20010db0)。不过,让我们看看以下内容:
struct netif {
...
u8_t hwaddr_len (at address 0x20010db8)
u8_t[8] hwaddr (at address 0x20010db9)
u8_t mtu (at address 0x20010dc1)
...
}
据我了解,hwaddr_len在4个字节上对齐,hwaddr在1个字节上“对齐”(因为它是u8_t,这不是在4个字节(32位)上对齐),mtu在1个字节上“对齐”。之后,结构体的所有其他成员再次对齐 4 个字节。所以,我认为这应该很好,即使 hwaddr 没有在 4 字节乘法器上对齐,但是当我尝试从“src”到 hwaddr 执行 memcpy 时,我收到了未对齐访问错误。
我正在arm gcc编译器上编译。有人知道为什么失败吗?
Ps:我对 ARM 对齐问题了解不多,抱歉,如果我的问题看起来很明显。
编辑:
编译器版本:gcc-arm-none-eabi-4_9-2015q3
失败的部分:
lpwif_get_slla(struct lpwif *lpwif, void *lla, unsigned char lla_len)
{
WpanDeviceP dev = lpwif->dev->wpan;
unsigned char len = 0;
if (lla_len >= 8) {
if (lpwif->eui[0] == 0xff) {
/* Fetch WPAN Device's Long Address. */
uint64_t addr64;
memset(&addr64, 0xff, sizeof(addr64));
WpanGet(dev, WpanPibAttr_macExtendedAddress, &addr64, 8);
/* Always return address in network-byte order */
lpwif->eui[0] = (addr64 >> 56) & 0xff;
lpwif->eui[1] = (addr64 >> 48) & 0xff;
lpwif->eui[2] = (addr64 >> 40) & 0xff;
lpwif->eui[3] = (addr64 >> 32) & 0xff;
lpwif->eui[4] = (addr64 >> 24) & 0xff;
lpwif->eui[5] = (addr64 >> 16) & 0xff;
lpwif->eui[6] = (addr64 >> 8) & 0xff;
lpwif->eui[7] = (addr64 >> 0) & 0xff;
}
if (lpwif->eui[0] == 0xff) return 0; /* Device has no EUI-64 address. */
if (lla) memcpy(lla, lpwif->eui, 8);
}
这个方法是通过 `lpwif_get_slla(&state->lpwif, netif->hwaddr, 8); 来调用的
拆解:
if (lpwif->eui[0] == 0xff) return 0; /* Device has no EUI-64 address. */
10098ac: 6bfb ldr r3, [r7, #60] ; 0x3c
10098ae: f893 3024 ldrb.w r3, [r3, #36] ; 0x24
10098b2: 2bff cmp r3, #255 ; 0xff
10098b4: d101 bne.n 10098ba <lpwif_get_slla+0x10e>
10098b6: 2300 movs r3, #0
10098b8: e07f b.n 10099ba <lpwif_get_slla+0x20e>
if (lla) memcpy(lla, lpwif->eui, 8);
10098ba: 6bbb ldr r3, [r7, #56] ; 0x38
10098bc: 2b00 cmp r3, #0
10098be: d006 beq.n 10098ce <lpwif_get_slla+0x122>
10098c0: 6bfb ldr r3, [r7, #60] ; 0x3c
10098c2: 3324 adds r3, #36 ; 0x24
10098c4: 6bb8 ldr r0, [r7, #56] ; 0x38
10098c6: 4619 mov r1, r3
10098c8: 2208 movs r2, #8
10098ca: 4b3f ldr r3, [pc, #252] ; (10099c8 <lpwif_get_slla+0x21c>)
10098cc: 4798 blx r3
return 8;
10098ce: 2308 movs r3, #8
10098d0: e073 b.n 10099ba <lpwif_get_slla+0x20e>
尝试使用简单的 for 循环将 memcpy 替换为副本。假设它是内存对齐的,编译器可能正在优化它。