alienware r4

; offset into the RAMAdd r0, R0, # phybase; add physical base // The final physical address of eboot is 0x30038000, Which is platform/smdk2440a/src/bootloader/eboot_usb/boot. the // Eboot link address: eboot 8c038000 00016800 ramimage // A. allows a ramimageworkflow to generate a corresponding. binfile, which is generated by romimage.exe. nb0 simultaneously fills the ptoc global variable pointer of eboot [Luther. gliethttp] MoV R1, R0; (R1) Copy destination// When the S3C2440 is started, the fir

lines of code, as shown below:LDR r0, = 0x7f008820LDR R1, = 0x1111STR R1, [R0]LDR r0, = 0x7f008824LDR R1, [R0]Bic R2, R1, #3ORR R2, R2, # 0xcBic R3, R1, # 0xcORR R3, R3, #3NN:STR R2, [R0]LDR R4, = 5000000Delay1:Sub R4, R4, #1CMP R4, #0BNE delay1STR R3, [R0]LDR R4, = 5000000

FunctionArea test_asm, code, readonlyExport test_asm_argsTest_asm_argsStr lr, [Sp, #-4]! ; Save the current LR.Stack is full decline stack FD, first adjust SP pointer, then press LR address.LDR r0, = 0x10; parameter 1LDR R1, = 0x20; parameter 2LDR R2, = 0x30; parameter 3LDR R3, = 0x40; parameter 4BL test_c_args; call the c FunctionLdr pc, [Sp], #4; load LR into PC (return main function), Pc = LR, SP = SP + 4, and restore the original stack.EndVoid test_c_args (int A, int B, int C, int D){Printk

CDP 1.On the first router (Router1), enter the terminal configuration mode. -Cl ick Here to View Answer-Router> enableRouter # conf tRouter (config )# 2.In Router1, change the nameR1 -Click Here to View Answer-Router (config) # hostname R1R1 (config )# 3.Modify the name of another router (Router4)R4 -Click Here to View Answer-Router> enableRouter # conf tRouter (config) # hostname R4R4 (config )# Note: by default, all ports are in the shutdown status

number Int s1/0 Ipv6 rip ABC enable must enter the RIP enabled under the interface Show ipv6 route Show ipv6 route rip OSPFV3 is used in IPV6 R1 (config) # Ipv6 router ospf 110 R1 (config-router) # Router-id 2.2.2.2Note::You must useIPV4Address ID,Must be manually specified,Cannot be automatically selected R1 (config) # Int s0 R1 (config-if) # Ipv6 ospf 110 area 0It is also advertised under the interface R1 (config) # Int lo0 R1 (config-if) # Ipv6 ospf 110 area 0The loopback port is still

OSPF Network failure handling cases Network topology 11.1 is shown. Network planning is as follows: N R1 and R2 's interconnected addresses are: 10.0.0.0/30;R2 and R3 's interconnected address: 20.0.0.0/30,R3 and R4 are connected to: 30.0.0.0/30;R4 and R1; n Use the Loopback0 interface address of the router as the loopback0:2.2.2.2/32,r3 of the ROUTE-ID,R1 loopback0:1.1.1.1/32,r2 loopback0:3.3.3.3/ The loop

point, you can see that when the scheduling occurs, the system is in a non-disruptive and the scheduler does not lock the case, call the Os_schednew function, the function part of the source code fory = osunmaptbl[osrdygrp];Ospriohighrdy = (int8u) ((y There are also some source code is the same function, only 63 priority scale to 255 priority, the process of finding a ready table changesOspriohighrdy represents the priority of the highest-priority tasks currently in the system, finds the highes

?There are four methods listed below, and the fourth method is the focus of today.1. We can realize the IBGP in as 234, but pay attention to the router configuration under the self-jumping, otherwise IBGP learned the route is not up to. Of course, this is a more cumbersome way, because the router, with command with the hand must be broken ah.2. We can also establish IBGP in R2 and R3, R3 and R4, establish a partial interconnect within as 234, and then

. Cisco also has a dedicated ppt to explain this problem, but I am still very vague at the end. Yun, but I remember a concept: the direction or interface of multicast traffic, show ip route x. x. x. the route table to which x is returned must be consistent. Otherwise, RPF fails and multicast traffic is discarded. Today, I found an experiment from my colleague. After completing the experiment, I found that RPF is not so mysterious. Here is the experiment process. Let's verify whether it is what I

. STRING, 0, 2, 1); this. testAddRows (rowMap2, 2, r2-2-3 (1), CellTypeEnum. STRING, 1, 2, 1); this. testAddRows (rowMap2, 1, r2-1-4 (1), CellTypeEnum. DIGIT, 2); rowMap2.put (rowtype, css2); // Add a HashMap rowMap3 = new HashMap (); lists. add (rowMap3); this. testAddRows (rowMap3, 1, r3-1-1 (1), CellTypeEnum. LABEL, 0, 0, 1); this. testAddRows (rowMap3, 1, r3-2-1 (1), CellTypeEnum. LABEL, 1, 0, 1); this. testAddRows (rowMap3, 1, r3-2-1 (1), CellTypeEnum. LABEL, 2, 0, 1); this. testAddRows (ro

SVC mode disable interruptions. MSR Cpsr_c, # (Os_cpu_arm_control_int_dis | OS_CPU_ARM_MODE_SVC); HANDLE NESTING Counter:ldr R3, __os_intnesting; osintnesting++; LDRB R4, [R3] ADD R4, R4, #1 STRB R4, [R3] MSR cpsr_cxsf, R2; RESTORE interrupted MO

jump command). For MIPS, it is saved in a register (that is, R31). For x86, it is saved to the stack (which will say what the stack is later). After the jump, execute a piece of code, you can also jump back, because the old PC saved. This is important if there are several places where you want to jump to this code and then go back to where it was. For example, the function of a piece of code is to do something based on the values of R4 and R5 (such a

The topology is Darip CCNP experiment Lab17, described as follows: R1 (S1/1)--r2 (s1/0), Cost=10;r1 (S1/2)-framerelay-r4 (S1/2), cost=2 R2 (S1/1)--R3 (s1/0), COST=2,R2 (f0/0)--r4 (f0/0), cost=2 R3 (S1/1)--r4 (s1/0), cost=5 R1 f0/0 Port Connection Network 192.168.1.0/24,cost=1 R1#show Run Interface fastethernet0/0IP address 192.168.1.1 255.255.255.0 ! Inte

R_ARM_ABS32.rel.dyn 0000b008 R_ARM_ABS32.rel.plt 0000AFDC r_arm_jump_slot In the code, we called 6 strlen, but why did it only occur 4 times? In addition, how do they correspond to each other, with these questions to analyze the assembly code. To drag the compiled so to Ida, we see the instructions for the sample code:. TEXT:000050BC EXPORT JAVA_COM_EXAMPLE_ALLHOOKINONE_HOOKUTILS_ELFHOOK.TEXT:000050BC Java_com_example_allho OKINONE_HOOKUTILS_ELFHOOK.TEXT:000050BC.TEXT:000050BC var_40 = -0X4

) # no shutdownR3-jin (config-if) # interface s1/0R3-jin (config-if) # ip address 192.168.1.1 255.255.255.0R3-jin (config-if) # no shutdownR3-jin (config-if) # exitR3-jin (config) # router ospf 1R3-jin (config-router) # network 172.16.0.1 0.0.0.0 area 0R3-jin (config-router) # network 192.168.1.1 0.0.0.0 area 2R3-jin (config-router) # endR3-jin #R3-jin # show ip ospf databaseR3-jin # show ip route3.0.0.0/32 is subnetted, 1 subnetsC 3.3.3.3 is directly connected, Loopback0O route 0.0.0/8 [110/65]

Concept preview: What is multi-point bidirectional routing re-release?Different routing protocols send route information to each other through re-publishing. The number of ASBR routers for re-publishing is more than one. in each routing protocol, the ASBR can receive the route information re-published by each other.I. Experiment topology: 650) this. width = 650; "alt =" "src =" http://www.bkjia.com/uploads/allimg/131227/0300461343-0.jpg "border =" 0 "/> Ii. Purpose:Dual-point: for the sake of r

[Experimental environment] liehuo.net C3640-IK9O3S-M Version 12.4 (10) liehuo.net [Tutorial Objective] veryhuo.com Uses Full-mesh to resolve BGP routing black hole liehuo.net [Experiment topology] liehuo.net Liehuo.net [Experiment description] lie # fire # Network BGP distribution. R2, R3, and R4 run the OSPF protocol. The objective is to enable 1.1.1.1 BGP neighbor relationship is established using the loopback interface. R1 Create a default route

ip address negotiated encapsulation PPP dialer Pool 1 NBSP;PPP chap hostname Cisco NBSP;PPP chap password 0 Cisco D.R4:Interface ethernet0/0IP address 192.168.1.4 255.255.255.0No shutdownIP Route 0.0.0.0 0.0.0.0 192.168.1.3E.R5:Interface ethernet0/0IP address 192.168.1.5 255.255.255.0No shutdownIP Route 0.0.0.0 0.0.0.0 192.168.1.3Four. Resolve Address Reflow---are all configured on R3.A. method one: NVI1. Dynamic PAT Configuration:IP Access-list Extended PATPermit IP 192.168.1.0 0.0.0.255

After early_init is executed in the previous article, the basic initialization work in the early stages of CPU startup is completed. Then the kernel starts to redirect and copy and run. The Code is as follows: Blreloc_offsetmrr26, r3addisr4, R3, kernelbase @ H/* Current address of _ start */lisr5, physical_start @ hcmplw0, R4, R5/* already running physical_start? */Bnerelocate_kernel/* Juan kernel redirection, required for classic startup */ Here, the

r0, = 0x7f008824 LDR R1, [R0] Bic R2, R1, #3 ORR R2, R2, # 0xc Bic R3, R1, # 0xc ORR R3, R3, #3 NN: STR R2, [R0] LDR R4, = 5000000 Delay1: Sub R4, R4, #1 CMP R4, #0 BNE delay1 STR R3, [R0] LDR R4, = 5000000 Delay2: Sub R4,