linux设备：cdev和kobj_map
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linux设备：cdev和kobj_map
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先看kobj_map相關(guān)的代碼
涉及到的文件
<linux/kobj_map.h>
<drivers/base/map.c>
[objc] view plaincopyprint?typedef?struct?kobject?*kobj_probe_t(dev_t,?intint?*,?voidvoid?*);?? struct?kobj_map;?? int?kobj_map(struct?kobj_map?*,?dev_t,?unsigned?long,?struct?module?*,?kobj_probe_t?*,?int?(*)(dev_t,?voidvoid?*),?voidvoid?*);?? void?kobj_unmap(struct?kobj_map?*,?dev_t,?unsigned?long);?? struct?kobject?*kobj_lookup(struct?kobj_map?*,?dev_t,?intint?*);?? struct?kobj_map?*kobj_map_init(kobj_probe_t?*,?struct?mutex?*);?? typedef struct kobject *kobj_probe_t(dev_t, int *, void *);
struct kobj_map;
int kobj_map(struct kobj_map *, dev_t, unsigned long, struct module *, kobj_probe_t *, int (*)(dev_t, void *), void *);
void kobj_unmap(struct kobj_map *, dev_t, unsigned long);
struct kobject *kobj_lookup(struct kobj_map *, dev_t, int *);
struct kobj_map *kobj_map_init(kobj_probe_t *, struct mutex *);
先看kobj_map結(jié)構(gòu)體
[objc] view plaincopyprint?struct?kobj_map?{?? ????struct?probe?{?? ????????struct?probe?*next;??????/*?這樣形成了鏈表結(jié)構(gòu)?*/?? ????????dev_t?dev;???????????????/*?設(shè)備號?*/?? ????????unsigned?long?range;?????/*?設(shè)備號的范圍?*/?? ????????struct?module?*owner;?? ????????kobj_probe_t?*get;?? ????????int?(*lock)?(dev_t,?voidvoid?*);?? ????????voidvoid?*data;??????????????/*?指向struct?cdev對象?*/?? ????}?*probes[255];?? ????struct?mutex?*lock;?? }?? struct kobj_map {struct probe {struct probe *next; /* 這樣形成了鏈表結(jié)構(gòu) */dev_t dev; /* 設(shè)備號 */unsigned long range; /* 設(shè)備號的范圍 */struct module *owner;kobj_probe_t *get;int (*lock) (dev_t, void *);void *data; /* 指向struct cdev對象 */} *probes[255];struct mutex *lock;
}結(jié)構(gòu)體中有一個互斥鎖lock，一個probes[255]數(shù)組，數(shù)組元素為struct probe的指針。根據(jù)下面的函數(shù)作用來看，kobj_map結(jié)構(gòu)體是用來管理設(shè)備號及其對應(yīng)的設(shè)備的。kobj_map函數(shù)就是將指定的設(shè)備號加入到該數(shù)組，kobj_lookup則查找該結(jié)構(gòu)體，然后返回對應(yīng)設(shè)備號的kobject對象，利用利用該kobject對象，我們可以得到包含它的對象如cdev。struct probe結(jié)構(gòu)體中的get函數(shù)指針就是用來獲得kobject對象的，可能不同類型的設(shè)備獲取的方式不同，我現(xiàn)在就看過cdev的exact_match函數(shù)。
kobj_map函數(shù)[objc] view plaincopyprint?int?kobj_map(struct?kobj_map?*domain,?dev_t?dev,?unsigned?long?range,?struct?module?*module,?kobj_probe_t?*probe,?int?(*lock)(dev_t,?voidvoid?*),?voidvoid?*data)?? {?? ????unsigned?n?=?MAJOR(dev+range-1)?-?MAJOR(dev)?+?1;?? ????unsigned?index?=?MAJOR(dev);?? ????unsigned?i;?? ????struct?probe?*p;?? ?? ????if?(n?>?255)?????/*?若n?>?255，則超出了kobj_map中probes數(shù)組的大小?*/?? ????????n?=?255;?? ????p?=?kmalloc(sizeof(struct?probe)?*?n,?GFP_KERNEL);??/*?分配n個struct?probe?*/?? ????if(p?==?NULL)?? ????????return?-ENOMEM;?? ????for(i?=?0;?i?<?n;?i++,?p++)?{?????/*?用函數(shù)的參數(shù)初始化probe?*/?? ????????p->owner?=?module;?? ????????p->get?=?probe;?? ????????p->lock?=?lock;?? ????????p->dev?=?dev;?? ????????p->range?=?range;?? ????????p->data?=?data;?? ????}?? ????mutex_lock(domain->lock);?? ????for(i?=?0,?p-=n;?i?<?n;?i++,?p++,?index++)?{?? ????????struct?probe?**s?=?&domain->probes[index?%?255];?? ????????while(*s?&&?(*s)->range?<?range)?? ????????????s?=?&(*s)->next;?? ????????p->next?=?*s;?? ????????*s?=?p;?? ????}?? ????mutex_unlock(domain->lock);?? ????return?0;?? }?? int kobj_map(struct kobj_map *domain, dev_t dev, unsigned long range, struct module *module, kobj_probe_t *probe, int (*lock)(dev_t, void *), void *data)
{unsigned n = MAJOR(dev+range-1) - MAJOR(dev) + 1;unsigned index = MAJOR(dev);unsigned i;struct probe *p;if (n > 255) /* 若n > 255，則超出了kobj_map中probes數(shù)組的大小 */n = 255;p = kmalloc(sizeof(struct probe) * n, GFP_KERNEL); /* 分配n個struct probe */if(p == NULL)return -ENOMEM;for(i = 0; i < n; i++, p++) { /* 用函數(shù)的參數(shù)初始化probe */p->owner = module;p->get = probe;p->lock = lock;p->dev = dev;p->range = range;p->data = data;}mutex_lock(domain->lock);for(i = 0, p-=n; i < n; i++, p++, index++) {struct probe **s = &domain->probes[index % 255];while(*s && (*s)->range < range)s = &(*s)->next;p->next = *s;*s = p;}mutex_unlock(domain->lock);return 0;
}
dev_t的前12位為主設(shè)備號，后20位為次設(shè)備號。n = MAJOR(dev + range - 1) - MAJOR(dev) + 1 表示設(shè)備號范圍(dev, dev+range)中不同的主設(shè)備號的個數(shù)。通常n的值為1。從代碼中的第二個for循環(huán)可以看出kobj_map中的probes數(shù)組中每個元素為一個struct probe鏈表的頭指針。每個鏈表中的probe對象有（MAJOR（probe.dev） % 255）值相同的關(guān)系。若主設(shè)備號小于255， 則每個鏈表中的probe都有相同的主設(shè)備號。鏈表中的元素是按照range值從小到大排列的。while循環(huán)即是找出該將p插入的位置。
kobj_unmap函數(shù)[objc] view plaincopyprint?void?kobj_unmap(struct?kobj_map?*domain,?dev_t?dev,?unsigned?long?range)?? {?? ????unsigned?n?=?MAJOR(dev?+?range?-?1)?-?MAJOR(dev)?+?1;?? ????unsigned?index?=?MAJOR(dev);?? ????unsigned?i;?? ????struct?probe?*found?=?NULL;?? ?? ????if?(n?>?255)?? ????????n?=?255;?? ?? ????mutex_lock(domain->lock);?? ????for?(i?=?0;?i?<?n;?i++,?index++)?{?? ????????struct?probe?**s;?? ????????for?(s?=?&domain->probes[index?%?255];?*s;?s?=?&(*s)->next)?{?? ????????????struct?probe?*p?=?*s;?? ????????????if?(p->dev?==?dev?&&?p->range?==?range)?{?? ????????????????*s?=?p->next;?? ????????????????if?(!found)?? ????????????????????found?=?p;?? ????????????????break;?? ????????????}?? ????????}?? ????}?? ????mutex_unlock(domain->lock);?? ????kfree(found);?? }?? void kobj_unmap(struct kobj_map *domain, dev_t dev, unsigned long range)
{unsigned n = MAJOR(dev + range - 1) - MAJOR(dev) + 1;unsigned index = MAJOR(dev);unsigned i;struct probe *found = NULL;if (n > 255)n = 255;mutex_lock(domain->lock);for (i = 0; i < n; i++, index++) {struct probe **s;for (s = &domain->probes[index % 255]; *s; s = &(*s)->next) {struct probe *p = *s;if (p->dev == dev && p->range == range) {*s = p->next;if (!found)found = p;break;}}}mutex_unlock(domain->lock);kfree(found);
}在16行，找到對應(yīng)設(shè)備號dev和range指定的probe對象后，退出，然后kfree釋放空間。
kobj_lookup函數(shù)[objc] view plaincopyprint?struct?kobject?*kobj_lookup(struct?kobj_map?*domain,?dev_t?dev,?intint?*index)?? {?? ????struct?kobject?*kobj;?? ????struct?probe?*p;?? ????unsigned?long?best?=?~0UL;?? ?? retry:?? ????mutex_lock(domain->lock);?? ????for?(p?=?domain->probes[MAJOR(dev)?%?255];?p;?p?=?p->next)?{?? ????????struct?kobject?*(*probe)(dev_t,?intint?*,?voidvoid?*);?? ????????struct?module?*owner;?? ????????voidvoid?*data;?? ?? ????????if?(p->dev?>?dev?||?p->dev?+?p->range?-?1?<?dev)?? ????????????continue;?? ????????if?(p->range?-?1?>=?best)?? ????????????break;?? ????????if?(!try_module_get(p->owner))?? ????????????continue;?? ????????owner?=?p->owner;?? ????????data?=?p->data;?? ????????probe?=?p->get;?? ????????best?=?p->range?-?1;?? ????????*index?=?dev?-?p->dev;???/*?這個是用來干嘛的？?*/?? ????????if?(p->lock?&&?p->lock(dev,?data)?<?0)?{?? ????????????module_put(owner);?? ????????????continue;?? ????????}?? ????????mutex_unlock(domain->lock);?? ????????kobj?=?probe(dev,?index,?data);?? ????????/*?Currently?->owner?protects?_only_?->probe()?itself.?*/?? ????????module_put(owner);?? ????????if?(kobj)?? ????????????return?kobj;?? ????????goto?retry;?? ????}?? ????mutex_unlock(domain->lock);?? ????return?NULL;?? }?? struct kobject *kobj_lookup(struct kobj_map *domain, dev_t dev, int *index)
{struct kobject *kobj;struct probe *p;unsigned long best = ~0UL;retry:mutex_lock(domain->lock);for (p = domain->probes[MAJOR(dev) % 255]; p; p = p->next) {struct kobject *(*probe)(dev_t, int *, void *);struct module *owner;void *data;if (p->dev > dev || p->dev + p->range - 1 < dev)continue;if (p->range - 1 >= best)break;if (!try_module_get(p->owner))continue;owner = p->owner;data = p->data;probe = p->get;best = p->range - 1;*index = dev - p->dev; /* 這個是用來干嘛的？ */if (p->lock && p->lock(dev, data) < 0) {module_put(owner);continue;}mutex_unlock(domain->lock);kobj = probe(dev, index, data);/* Currently ->owner protects _only_ ->probe() itself. */module_put(owner);if (kobj)return kobj;goto retry;}mutex_unlock(domain->lock);return NULL;
}
對cdev_add函數(shù)，這里的p->probe函數(shù)即是exact_match, p->lock為exact_lock函數(shù)。
kobj_map_init函數(shù)[objc] view plaincopyprint?struct?kobj_map?*kobj_map_init(kobj_probe_t?*base_probe,?struct?mutex?*lock)?? {?? ????struct?kobj_map?*p?=?kmalloc(sizeof(struct?kobj_map),?GFP_KERNEL);?? ????struct?probe?*base?=?kzalloc(sizeof(*base),?GFP_KERNEL);?? ????int?i;?? ?? ????if?((p?==?NULL)?||?(base?==?NULL))?{?? ????????kfree(p);?? ????????kfree(base);?? ????????return?NULL;?? ????}?? ?? ????base->dev?=?1;?? ????base->range?=?~0;?? ????base->get?=?base_probe;?? ????for?(i?=?0;?i?<?255;?i++)?? ????????p->probes[i]?=?base;?? ????p->lock?=?lock;?? ????return?p;?? }?? struct kobj_map *kobj_map_init(kobj_probe_t *base_probe, struct mutex *lock)
{struct kobj_map *p = kmalloc(sizeof(struct kobj_map), GFP_KERNEL);struct probe *base = kzalloc(sizeof(*base), GFP_KERNEL);int i;if ((p == NULL) || (base == NULL)) {kfree(p);kfree(base);return NULL;}base->dev = 1;base->range = ~0;base->get = base_probe;for (i = 0; i < 255; i++)p->probes[i] = base;p->lock = lock;return p;
}
在初始化一個kobj_map對象時，將probes指針全部指向同一個base。
下面是cdev部分。文件：<linux/cdev.h><fs/char_dev.c>
cdev.h[objc] view plaincopyprint?struct?cdev?{?? ????struct?kobject?kobj;?? ????struct?module?*owner;?? ????const?struct?file_operations?*ops;?? ????struct?list_head?list;?? ????dev_t?dev;?? ????unsigned?int?count;?? }?? void?cdev_init(struct?cdev?*,?const?struct?file_operations?*);?? struct?cdev?*cdev_alloc(void);?? void?cdev_put(struct?cdev?*p);?? int?cdev_add(struct?cdev?*,?dev_t,?unsigned);?? void?cdev_del(struct?cdev?*);?? struct cdev {struct kobject kobj;struct module *owner;const struct file_operations *ops;struct list_head list;dev_t dev;unsigned int count;
}
void cdev_init(struct cdev *, const struct file_operations *);
struct cdev *cdev_alloc(void);
void cdev_put(struct cdev *p);
int cdev_add(struct cdev *, dev_t, unsigned);
void cdev_del(struct cdev *);
cdev_init函數(shù)此函數(shù)首先調(diào)用kobject_init初始化cdev中的kobj，然后將cdev中的ops賦值。
cdev_alloc函數(shù)先kzalloc分配一個cdev，然后用kobject_init初始化kobj
cdev_put函數(shù)
[objc] view plaincopyprint?void?cdev_put(struct?cdev?*p)?? {?? ????if?(p)?{?? ????????struct?module?*owner?=?p->owner;?? ????????kobject_put(&p->kobj);?? ????????module_put(owner);?? ????}?? }?? void cdev_put(struct cdev *p)
{if (p) {struct module *owner = p->owner;kobject_put(&p->kobj);module_put(owner);}
}
此函數(shù)調(diào)用kobject_put和module_put，好像它們的作用就是減少引用計數(shù)
cdev_add函數(shù)[objc] view plaincopyprint?int?cdev_add(struct?cdev?*p,?dev_t?dev,?unsigned?count)?? {?? ????p->dev?=?dev;?? ????p->count?=?count;?? ????return?kobj_map(cdev_map,?dev,?count,?NULL,?exact_match,?exact_lock,?p);?? }?? int cdev_add(struct cdev *p, dev_t dev, unsigned count)
{p->dev = dev;p->count = count;return kobj_map(cdev_map, dev, count, NULL, exact_match, exact_lock, p);
}
主要是調(diào)用kobj_map將cdev放入cdev_map中。
cdev_del函數(shù)[objc] view plaincopyprint?static?void?cdev_unmap(dev_t?dev,?unsigned?count)?? {?? ????kobj_unmap(cdev_map,?dev,?count);?? }?? ?? void?cdev_del(struct?cdev?*p)?? {?? ????cdev_unmap(p->dev,?p->count);?? ????kobject_put(&p->kobj);?? }?? static void cdev_unmap(dev_t dev, unsigned count)
{kobj_unmap(cdev_map, dev, count);
}void cdev_del(struct cdev *p)
{cdev_unmap(p->dev, p->count);kobject_put(&p->kobj);
}
這就不用說啥了。
LDD3上說“只要cdev_add返回了，我們的設(shè)備就‘活’了，它的操作就會被內(nèi)核調(diào)用"，那么這句奇妙的話到底是個什么意思？
下面是我目前了解的情況
據(jù)說在open一個字符設(shè)備文件時，最終總會調(diào)用chrdev_open。下面是該函數(shù)的源碼注意inode->i_rdev中保存了設(shè)備編號，inode->icdev指向了cdev結(jié)構(gòu)。[objc] view plaincopyprint?static?int?chrdev_open(struct?inode?*inode,?struct?file?*filp)?? {?? ????struct?cdev?*p;?? ????struct?cdev?*new?=?NULL;?? ????int?ret?=?0;?? ?? ????spin_lock(&cdev_lock);?? ????p?=?inode->i_cdev;?? ????if?(!p)?{?? ????????struct?kobject?*kobj;?? ????????int?idx;?? ????????spin_unlock(&cdev_lock);?? ????????kobj?=?kobj_lookup(cdev_map,?inode->i_rdev,?&idx);????? ????????if?(!kobj)?? ????????????return?-ENXIO;?? ????????new?=?container_of(kobj,?struct?cdev,?kobj);???/*?找到字符設(shè)備的cdev?*/?? ????????spin_lock(&cdev_lock);?? ????????/*?Check?i_cdev?again?in?case?somebody?beat?us?to?it?while? ?????????we?dropped?the?lock.?*/?? ????????p?=?inode->i_cdev;?? ????????if?(!p)?{?? ????????????inode->i_cdev?=?p?=?new;?? ????????????list_add(&inode->i_devices,?&p->list);/*?ZXG:?這是啥？?*/?? ????????????new?=?NULL;?? ????????}?else?if?(!cdev_get(p))?? ????????????ret?=?-ENXIO;?? ????}?else?if?(!cdev_get(p))?? ????????ret?=?-ENXIO;?? ????spin_unlock(&cdev_lock);?? ????cdev_put(new);?? ????if?(ret)?? ????????return?ret;?? ?? ????ret?=?-ENXIO;?? ????filp->f_op?=?fops_get(p->ops);?? ????if?(!filp->f_op)?? ????????goto?out_cdev_put;?? ?? ????if?(filp->f_op->open)?{?? ????????ret?=?filp->f_op->open(inode,?filp);?/*?調(diào)用cdev->ops中的open函數(shù)?*/?? ????????if?(ret)?? ????????????goto?out_cdev_put;?? ????}?? ?? ????return?0;?? ?? ?out_cdev_put:?? ????cdev_put(p);?? ????return?ret;?? }?? static int chrdev_open(struct inode *inode, struct file *filp)
{struct cdev *p;struct cdev *new = NULL;int ret = 0;spin_lock(&cdev_lock);p = inode->i_cdev;if (!p) {struct kobject *kobj;int idx;spin_unlock(&cdev_lock);kobj = kobj_lookup(cdev_map, inode->i_rdev, &idx); if (!kobj)return -ENXIO;new = container_of(kobj, struct cdev, kobj); /* 找到字符設(shè)備的cdev */spin_lock(&cdev_lock);/* Check i_cdev again in case somebody beat us to it whilewe dropped the lock. */p = inode->i_cdev;if (!p) {inode->i_cdev = p = new;list_add(&inode->i_devices, &p->list);/* ZXG: 這是啥？ */new = NULL;} else if (!cdev_get(p))ret = -ENXIO;} else if (!cdev_get(p))ret = -ENXIO;spin_unlock(&cdev_lock);cdev_put(new);if (ret)return ret;ret = -ENXIO;filp->f_op = fops_get(p->ops);if (!filp->f_op)goto out_cdev_put;if (filp->f_op->open) {ret = filp->f_op->open(inode, filp); /* 調(diào)用cdev->ops中的open函數(shù) */if (ret)goto out_cdev_put;}return 0;out_cdev_put:cdev_put(p);return ret;
}
先看kobj_map結(jié)構(gòu)體
[objc] view plaincopyprint?
kobj_map函數(shù)[objc] view plaincopyprint?
dev_t的前12位為主設(shè)備號，后20位為次設(shè)備號。n = MAJOR(dev + range - 1) - MAJOR(dev) + 1 表示設(shè)備號范圍(dev, dev+range)中不同的主設(shè)備號的個數(shù)。通常n的值為1。從代碼中的第二個for循環(huán)可以看出kobj_map中的probes數(shù)組中每個元素為一個struct probe鏈表的頭指針。每個鏈表中的probe對象有（MAJOR（probe.dev） % 255）值相同的關(guān)系。若主設(shè)備號小于255， 則每個鏈表中的probe都有相同的主設(shè)備號。鏈表中的元素是按照range值從小到大排列的。while循環(huán)即是找出該將p插入的位置。
kobj_unmap函數(shù)[objc] view plaincopyprint?
kobj_lookup函數(shù)[objc] view plaincopyprint?
對cdev_add函數(shù)，這里的p->probe函數(shù)即是exact_match, p->lock為exact_lock函數(shù)。
kobj_map_init函數(shù)[objc] view plaincopyprint?
在初始化一個kobj_map對象時，將probes指針全部指向同一個base。
下面是cdev部分。文件：<linux/cdev.h><fs/char_dev.c>
cdev.h[objc] view plaincopyprint?
cdev_init函數(shù)此函數(shù)首先調(diào)用kobject_init初始化cdev中的kobj，然后將cdev中的ops賦值。
cdev_alloc函數(shù)先kzalloc分配一個cdev，然后用kobject_init初始化kobj
cdev_put函數(shù)
[objc] view plaincopyprint?
此函數(shù)調(diào)用kobject_put和module_put，好像它們的作用就是減少引用計數(shù)
cdev_add函數(shù)[objc] view plaincopyprint?
主要是調(diào)用kobj_map將cdev放入cdev_map中。
cdev_del函數(shù)[objc] view plaincopyprint?
這就不用說啥了。
LDD3上說“只要cdev_add返回了，我們的設(shè)備就‘活’了，它的操作就會被內(nèi)核調(diào)用"，那么這句奇妙的話到底是個什么意思？
下面是我目前了解的情況
據(jù)說在open一個字符設(shè)備文件時，最終總會調(diào)用chrdev_open。下面是該函數(shù)的源碼注意inode->i_rdev中保存了設(shè)備編號，inode->icdev指向了cdev結(jié)構(gòu)。[objc] view plaincopyprint?
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