函数逻辑报告

函数名称：zbud_free() - frees the allocation associated with the given handle*@pool: pool in which the allocation resided*@handle: handle associated with the allocation returned by zbud_alloc()* In the case that the zbud page in which the allocation resides is

函数原型：void zbud_free(struct zbud_pool *pool, unsigned long handle)

返回类型：void

参数：