SDIO驱动(14)card的CIS读取及解析

关于CIS需要清楚的:

1、CIS是什么

CIS-Card Information Structure的缩写,CIS的作用:

The CIS includes information on power, function, manufacturer and other things the host
needs to determine if the I/O function(s) is appropriate to power-up.

2、CIS的呈现

SDIO驱动(14)card的CIS读取及解析_第1张图片
CIS位于CIA区域,地址寻址范围0x001000~0x017FFF。CIS有两类:

(1)Common CIS,记录整张card的通用信息,比如厂商、VID等信息。

(2)Function CIS,每一个Function的特有信息。

一个CIS元素的组织结构:

SDIO驱动(14)card的CIS读取及解析_第2张图片

它由结构体sdio_func_tuple表征:

/*
 * SDIO function CIS tuple (unknown to the core)
 */
struct sdio_func_tuple {
	struct sdio_func_tuple *next;
	unsigned char code;
	unsigned char size;
	unsigned char data[0];
};

这里总结下Function:Function代表card的一个具体功能,每一个card具有多个Function,Function的标号为0~7,特别地,0号针对的是CIA(可以简单理解就是用来访问CIA区域的)。在软件层面,Function由struct sdio_func表示,它作为一个device注册到系统中。

3、CIS的访问

需要两个条件:命令CMD52;CIS地址。

对于Common CIS,其地址保存在CCCR的0x09~0x0B地址处的寄存器中,3个寄存器值组成一个24位的地址。

对于Function CIS,其地址保存在每一个Function的FBR(Function Basic Registers)地址0x109-0x10B中。

软件层面,通过sdio_read_cis函数读取,函数原型:

static int sdio_read_cis(struct mmc_card *card, struct sdio_func *func)
如果参数func为NULL,访问的是Common CIS;否则则为指定Function的CIS。
4、CIS的解析

这正是接下来的重点要说的。

static int sdio_read_cis(struct mmc_card *card, struct sdio_func *func)
{
	int ret;
	struct sdio_func_tuple *this, **prev;
	unsigned i, ptr = 0;

	/*
	 * Note that this works for the common CIS (function number 0) as
	 * well as a function's CIS * since SDIO_CCCR_CIS and SDIO_FBR_CIS
	 * have the same offset.
	 */
	for (i = 0; i < 3; i++) {
		unsigned char x, fn;

		if (func)
			fn = func->num;
		else
			fn = 0;

		ret = mmc_io_rw_direct(card, 0, 0,
			SDIO_FBR_BASE(fn) + SDIO_FBR_CIS + i, 0, &x);
		if (ret)
			return ret;
		ptr |= x << (i * 8);
	}

	if (func)
		prev = &func->tuples;
	else
		prev = &card->tuples;

	BUG_ON(*prev);

	do {
		unsigned char tpl_code, tpl_link;

		ret = mmc_io_rw_direct(card, 0, 0, ptr++, 0, &tpl_code);
		if (ret)
			break;

		/* 0xff means we're done */
		if (tpl_code == 0xff)
			break;

		/* null entries have no link field or data */
		if (tpl_code == 0x00)
			continue;

		ret = mmc_io_rw_direct(card, 0, 0, ptr++, 0, &tpl_link);
		if (ret)
			break;

		/* a size of 0xff also means we're done */
		if (tpl_link == 0xff)
			break;

		this = kmalloc(sizeof(*this) + tpl_link, GFP_KERNEL);
		if (!this)
			return -ENOMEM;

		for (i = 0; i < tpl_link; i++) {
			ret = mmc_io_rw_direct(card, 0, 0,
					       ptr + i, 0, &this->data[i]);
			if (ret)
				break;
		}
		if (ret) {
			kfree(this);
			break;
		}

		/* Try to parse the CIS tuple */
		ret = cis_tpl_parse(card, func, "CIS",
				    cis_tpl_list, ARRAY_SIZE(cis_tpl_list),
				    tpl_code, this->data, tpl_link);
		if (ret == -EILSEQ || ret == -ENOENT) {
			/*
			 * The tuple is unknown or known but not parsed.
			 * Queue the tuple for the function driver.
			 */
			this->next = NULL;
			this->code = tpl_code;
			this->size = tpl_link;
			*prev = this;
			prev = &this->next;

			if (ret == -ENOENT) {
				/* warn about unknown tuples */
				printk(KERN_WARNING "%s: queuing unknown"
				       " CIS tuple 0x%02x (%u bytes)\n",
				       mmc_hostname(card->host),
				       tpl_code, tpl_link);
			}

			/* keep on analyzing tuples */
			ret = 0;
		} else {
			/*
			 * We don't need the tuple anymore if it was
			 * successfully parsed by the SDIO core or if it is
			 * not going to be queued for a driver.
			 */
			kfree(this);
		}

		ptr += tpl_link;
	} while (!ret);

	/*
	 * Link in all unknown tuples found in the common CIS so that
	 * drivers don't have to go digging in two places.
	 */
	if (func)
		*prev = card->tuples;

	return ret;
}
12~25行,获取CIS的地址。保存信息的寄存器地址计算方式:SDIO_FBR_BASE(fn) + SDIO_FBR_CIS + i = fn * 0x100 + 0x09 + i,由于地址排列为little-endian格式(LSB first)所以依次左移拼成一个24位地址。
27~30行,之前说过,Common CIS是属于整张card的(30行),而Function CIS是属于某个Function的(28行)。由于CIS信息都为设置,所以*prev指向的内容应该为NULL(32行)。

34~107行的代码段即是该函数的核心。结合上面提到的CIS的组织结构,当前ptr指向的一个CIS的code号码。

42行,code为0xFF标识一个CIS的结束。

46行,code为0x00标识当前元组(tuple)没什么内容,continue下一个tuple。

49行,当前ptr指向link域,标识当前tuple的数据size,同时也标识下一个tuple的偏移offset。

54行,link为0xFF同样标识一个CIS的结束。

57~70行,当前ptr指向数据区域的开始,读取数据;73~75行,解析数据:

static int cis_tpl_parse(struct mmc_card *card, struct sdio_func *func,
			 const char *tpl_descr,
			 const struct cis_tpl *tpl, int tpl_count,
			 unsigned char code,
			 const unsigned char *buf, unsigned size)
{
	int i, ret;

	/* look for a matching code in the table */
	for (i = 0; i < tpl_count; i++, tpl++) {
		if (tpl->code == code)
			break;
	}
	if (i < tpl_count) {
		if (size >= tpl->min_size) {
			if (tpl->parse)
				ret = tpl->parse(card, func, buf, size);
			else
				ret = -EILSEQ;	/* known tuple, not parsed */
		} else {
			/* invalid tuple */
			ret = -EINVAL;
		}
		if (ret && ret != -EILSEQ && ret != -ENOENT) {
			printk(KERN_ERR "%s: bad %s tuple 0x%02x (%u bytes)\n",
			       mmc_hostname(card->host), tpl_descr, code, size);
		}
	} else {
		/* unknown tuple */
		ret = -ENOENT;
	}

	return ret;
}
参数tpl、tpl_count为自己定义的一个cis_tpl类型数组及数组元素个数:
typedef int (tpl_parse_t)(struct mmc_card *, struct sdio_func *,
			   const unsigned char *, unsigned);
struct cis_tpl {
	unsigned char code;
	unsigned char min_size;
	tpl_parse_t *parse;
};

/* Known TPL_CODEs table for CIS tuples */
static const struct cis_tpl cis_tpl_list[] = {
	{	0x15,	3,	cistpl_vers_1		},
	{	0x20,	4,	cistpl_manfid		},
	{	0x21,	2,	/* cistpl_funcid */	},
	{	0x22,	0,	cistpl_funce		},
};
cis_tpl的作用是,根据不同的code调用不同的数据解析函数parse。
cis_tpl_parse函数根据传参code进行比对,找出处理该code的解析函数,然后调用之完成解析。以code=20H为例,调用函数cistpl_manfid解析数据:

static int cistpl_manfid(struct mmc_card *card, struct sdio_func *func,
			 const unsigned char *buf, unsigned size)
{
	unsigned int vendor, device;

	/* TPLMID_MANF */
	vendor = buf[0] | (buf[1] << 8);

	/* TPLMID_CARD */
	device = buf[2] | (buf[3] << 8);

	if (func) {
		func->vendor = vendor;
		func->device = device;
	} else {
		card->cis.vendor = vendor;
		card->cis.device = device;
	}

	return 0;
}
SDIO驱动(14)card的CIS读取及解析_第3张图片
返回到sdio_read_cis函数,76~104行如果解析成功(else分支)那么信息已经保存到相应的card或func中,所以可以释放掉之前malloc的内存;否则就把该信息保存到card或func的tuples指向的内存区域。

106行,ptr指向下一个tuple的code位置,开始下一轮读取、解析。

113、114行,把所有未知的tuple统一放到card的tuples里面。

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