timberland pro boots rough understanding of linux

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Linux to based I / O mapping of the I / O port and memory mapping approach based on I / O port resources, collectively referred to as I / O Region). I / O Region is still an I / O resources, it can still be used to describe the type of resource structure.
Linux is an inverted tree structure to manage each type of I / O resources (such as: I / O ports, peripheral memory, DMA and IRQ) for. each type of I / O resources are the resources corresponding to have an inverted tree,herve leger style dress, each node in the tree is a resource structure, and tree the root of the root of such resources is described in the resource space.
1. structure
1.1> struct resource iomem_resource = {1.2> struct resource {
const char * name;
unsigned long start, end;
unsigned long flags;
struct resource * parent, * sibling, * child;
2. Call function
request_mem_region (S1D_PHYSICAL_REG_ADDR, S1D_PHYSICAL_REG_SIZE,tory burch flats sale, Check for safe occupancy after the start physical address S1D_PHYSICAL_REG_ADDR S1D_PHYSICAL_REG_SIZE consecutive bytes of space struct resource * __request_region (struct resource * parent, unsigned long start, unsigned long n, const char * name) struct resource * res = kmalloc (sizeof ( * res), GFP_KERNEL); if (res) {memset (res, 0, sizeof (* res)); res-> name = name; res-> start = start; res-> end = start + n - 1; res-> flags = IORESOURCE_BUSY; write_lock (& ​​resource_lock); for (;;) {struct resource * conflict; conflict = __request_resource (parent, res); / / sibling parent under all unit test for partial the existence of overlapping conflicts if (! conflict) / / conflict = 0; application is successful, the normal placement of the [start, end] to the corresponding position break; if (conflict! = parent) {parent = conflict; if (! (conflict- > flags & IORESOURCE_BUSY)) continue; kfree (res); / / resource overlap conflict is detected, kfree application for the return kmalloc memory res = NULL; break; write_unlock (& ​​resource_lock); return res; static struct resource * __request_resource (struct resource * root, struct resource * new) unsigned long start = new-> start; unsigned long end = new-> end; struct resource * tmp, ** p; if (end start) return root; if (end> root-> end) return root; p = & root-> child; / / root element under the first list * p . [child list is based on I / O resources, physical address from low to high in the order of] for (;;) {tmp = * p; if (! tmp | | tmp-> start> end) {new- > sibling = tmp; * p = new; / / from root-> child = null begin our analysis, then tmp = null, then the first applications will be! tmp conditions are met, and enter / / this time root-> child the value of new indicators, new-> sibling = tmp = null; When the second application occurs: If the tmp-> start> end set up, / / ​​then, root-> child's new pointer value, new-> sibling = tmp; This link, and spatial distribution, such as: / / child = [start, end ]-->[ tmp-> start, tmp-> end] (1); If the conditions tmp-> start> end does not hold, then only! tmp conditions of access / / Well, root-> child the value of the constant, tmp-> sibling = new; new-> sibling = tmp = null This link, and spatial distribution, such as: / / child = [child-> start, child-> end ]-->[ start, end] (2); / / When the first time of three applications: If you start in (2) of the [child-> end, end], between then tmp-> end start slibing = null, / / ​​because the tmp-> end> ​​start, so the resource conflict,ladies rolex watches for sale, return ( 2) the [start, end] domain / / sum of the two boundary conditions and a median value of the analysis, we can see a code to address to high address from the order of list / / model diagram: childe = [a, b ]-->[ c,南非女子与11只猛兽同吃同睡(图), d ]-->[ e, f], then there is a [x, y] to be inserted into, tmp as a sibling pointer swimming / / tmp point to the child = [ ,],[> [c, d ]-->[ e, f] -> null; when tmp-> end> ​​= x, the conflict to return tmp / / tmp point to [c, d],affordable rolex watches Listen recipe for medicine man swallowed two bowls of l, if tmp-> ; start> y, insert the link after the photo: child = [a, b ]-->[ x, y ]-->[ c, d ]-->[ e, f] -> ; null; when tmp-> end> ​​= x,timberland pro boots, the conflict to return tmp / / tmp point to [e, f], if tmp-> start> y, insert the link after the photo: child = [a, b ]-->[ c, d ]-->[ x, y ]-->[ e,ed hardy hoody, f] -> null; when tmp-> end> ​​= x, the conflict to return tmp / / tmp point to null,supra vaider 小伙为示爱在斑马线上做俯卧撑(图), insert the link after the photo: child = [a, b ]-->[ c, d ]-->[ e, f ]-->[ x, y] -> null ; / / smooth to the detection of conflict, the purpose of the order link new-> parent = root; return NULL; p = & tmp-> sibling; if (tmp-> end <start) continue; return tmp ; article from CSDN blog, reproduced, please credit: http://
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