r5 spotify

; Add. s assembler stack_top equ 0x40002000preserve8export addarea add, code, readonlyimport sumexport startentrystartldr R13, = stack_top mov r0, # 1mov R1, # 100bl sumldr R5, = resultstr r0, [R5] resultdcd 0end // sum. C program int sum (int A, int B) {int sum = 0, I; for (I = A; I Assignment 4 1. Area myarea, code entry mov r0, #9 mov R1, #15 startcmp r0, r1bne judgebeq stopsub1sub r0, R0, r1sub2sub

in the MPLS domain, from one LSP to two LSP. use router C as the demarcation point. non-link aggregation, a complete FEC path: In the following topology, R4 has a loopback port. 44.1.1.1/32R5 also has a loopback port: During the 55.1.1.1/32 period, R4, R1, R2, R3, R5, all devices have the same understanding of the two routes (with the same mask, to form a prerequisite for completing the FEC link. from R4 to R 5 is a path, which can be viewed through

650) this. width = 650; "src =" http://www.bkjia.com/uploads/allimg/131227/055614D55-0.png "title =" cisco.png "/> 650) this. width = 650; "src =" http://www.bkjia.com/uploads/allimg/131227/0556141032-1.png "title =" topological configuration .png "/> Basic topology of an enterprise: Idea: 1) configure the aggregation layer and access layer first, and then configure the core layer and route 2) SW5 SW6 SW7 SW8 configure the port trunk mode, bind the port, and create vtp. 3) configure the vlan ad

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] area 2 [R4-ospf-1-area-0.0.0.2] network 192.168.4.2 0.0.0.0 [R4-ospf-1-area-0.0.0.2] network 192.168.5.1 0.0.0.0 Ar5 [Huawei] sysname r5 [R5] int s0/0/0 [R5-Serial0/0/0] ip add 192.168.6.2 24 [R5-Serial0/0/0] int s0/0/1 [R5-Serial0/0/1] ip add 192.168.7.1 24 [

be seen from the figure, only 4 DLCI is allowed, so it needs to be configured as Hub-spoke mode. Where R4 is the hub R2,R3,R4 's physical interface connects Frame-relay SWITCH, Between R4 and R5. Please use a PPP link. Eliminate redundant 32-bit host routing PPP Link PPP pushes its own host routing, is to ensure that there is no address on the interface to borrow the back of the network segment address, push to the other

=" consider using RAID 50 to replace raid ten (SSD) "style=" Border:0px;vertical-align:middle;height:auto; />RAID 50 in the above comparison has been able to tolerate the 3rd or even 4th disk failure, but the availability is lower than raid 10, but neither of them can achieve the perfect 100%, so the tradeoff between availability and cost of RAID 50 is a considerable advantage.Next look at performance, in order to be able to analyze performance well, we followed the first set of comparison scena

PGD = c064c000[00000000] * PGD = e0010031, * PTE = 00000000, * PPTE = 00000000Internal error: Oops: 817 [#4]Modules linked in: rt73 hiwdt protocol NFS lockd SunRPC mmc_block himci mmc_core protocol used hiusb serial numbers hifb FB serial numbers too many serial numbers exceed hi3511_group serial numbers exceed hi3511_base TDE protocol stkmediactl serial numbers exceed hi3511_base gpioi2c hiether mmzCPU: 0PC is at _ queue_work + 0x48/0x68LR is at 0xc03254f0PC: [SP: c046fb10 IP: c042c848 FP: c046

From: feino Network (www.firnow.com): http://dev.firnow.com/course/6_system/linux/Linuxjs/20091016/179018.html The below is a example of kernel panic log. [2, 875.497504] unable to handle kernel paging request at virtual address 4042a00c[1, 875.499110] PGD = c0004000[875.501797] [4042a00c] * PGD = 00000000[2, 875.505364] internal error: Oops: 5 [#1] preempt[875.509957] modules linked in: DHD[1, 875.513350] CPU: 0 not tainted (2.6.29 #21)[875.517877] PC is at lock_acquire + 0x60/0x74[875.522024]

: [ ] (Initi_irq_demux_extint4t7 + 0x0/0xa8) from [ ] (Asm_do_irq + 0x 44/0 x5c) R4: c03c2350 [ ] (Asm_do_irq + 0x0/0x5c) from [ ] (_ Irq_svc + 0x24/0 xa0) Exception stack (0xc2e2bd98 to 0xc2e2bde0) Bd80: 00000000 fb000000 Bda0: 00000001 00000000 c03c2a18 00000033 40000013 00000080 00000033 Bdc0: c39bfc80 c2e2be10 c2e2bdbc c2e2bde0 c007320c c0072748 60000013 ffffffff R7: 00000080 R6: 00000010 R5: f4000000 R4: ffffffff

code to the beginning of the probe function:int *ptest = NULL;*ptest = 0x1234;Re-compile the kernel, and it will error after startup and print out the following Oops information:Unable to handle kernel NULL pointer dereference in virtual Address 00000000PGD = c0004000[00000000] *pgd=00000000Internal error:oops:805 [#1]Modules Linked In:cpu:0Not tainted (2.6.22.6 #36)PC is at s3c2410fb_probe+0x18/0x560LR is at platform_drv_probe+0x20/0x24PC: [LR: [psr:a0000013Sp:c0481e64 ip:c0481ea0 fp:c0481e9cr

(config) # int lo0R3 (config-If) # IP add 3.3.3.3 255.255.255.0R3 (config) # int S1/2R3 (config-If) # encapsulation frame-relayR3 (config-If) # No frame-relay inverse-ARPR3 (config-If) # No ARP frame-relayR3 (config-If) # IP add 123.1.1.3 255.255.255.0R3 (config-If) # frame-relay map IP 123.1.1.2 302 Broadcast Certificate ---------------------------------------------------------------------------------------------------------------------------------------- R4 (config) # int lo0R4 (config-If) #

EL5_U8 i386:http://ftp.heanet.ie/disk1/oracle-linux/EL5/U8/i386/OracleLinux-R5-U8-Server-i386-dvd.isox86_64:http://ftp.heanet.ie/disk1/oracle-linux/EL5/U8/x86_64/OracleLinux-R5-U8-Server-x86_64-dvd.iso EL5_U9 i386:http://ftp.heanet.ie/disk1/oracle-linux/EL5/U9/i386/Enterprise-R5-U9-Server-i386-dvd.isox86_64:http://ftp.heanet.ie/disk1/oracle-linux/EL5/U9/x86_

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

???? R0, R0, IP, LSL r04:???? 00000002???? ANDEQ???? R0, R0, R28:???? 00040000???? ANDEQ???? R0, R4, R0C:???? 00000000???? ANDEQ???? R0, R0, R0Ten:???? 50008000???? ANDPL???? R8, R0, R0:???? 0000000c???? ANDEQ???? R0, R0, IP???? ...Disassembly of section. Debug_info:?00000000 0:???? 00000045???? ANDEQ???? R0, R0, R5, ASR #324:???? 00000002???? ANDEQ???? R0, R0, R28:???? 01040000???? Tsteq???? R4, R0C:???? 00000000???? ANDEQ???? R0, R0, R0Ten:???? 5000

1. How to determine whether a basic amplification circuit is a common-base, co-injection or a common set? Remove the input, the output side, the rest is the common end. The following figure: The UI is connected to the base pole, and the UO is the common collector. 2. Basic Amplification Circuit: 2-1. Co-emitting minimal signal amplifier Function: Amplify the small amplitude voltage signal Working principle: Q1 is connected to the common-injection configuration single-tube amplifier, R1,R2 pr

the three stages of the first 22 stages.For a more detailed description of the match, see an earlier article Nabila Amarsy.Phase 1: Problem solution matchingAt this stage, you don't have enough evidence to prove that your client really cares about the value proposition of your product. This writing idea is still in writing, or just on paper.Apple, for example, is "written" convinced that their Apple Music streaming services will provide more value to customers than

= 000000a0 R02 = 08000100 R03 = 08008000R04 = 00000000 R05 = 08000124 R06 = 083dc0a9 R07 = 0841bc9cR08 = 083DBFDC R09 = 083E0260 R10 (SL) = 00000000 R11 (FP) = 00000002R12 (IP) = 083dbfc0 R13 (SP) = 083dbd44 R14 (LR) = 08413984 PC = 08008000CPSR = 600000d3 SPSR = b00000ff>bkm>di 0x08008000//disassemble the address area code08008000:mov r0,r008008004:mov r0,r008008008:mov r0,r00800800c:mov r0,r008008010:mov r0,r008008014:mov r0,r008008018:mov r0,r00800801c:mov r0,r008008020:b 0x800803008008024:.

abstract, and here is an example to illustrate briefly, it is easy to understand:Arm-inux-objdump–d u-boot > Dump_u-boot.txtYou can get the Dump_u-boot.txt file. The file is in, containing the executable assembly code of the program in U-boot, where we can see the source code of the C language function, which corresponds to those assembly code.The following is a compilation code of two functions,One is Clock_init,The other is not clock_init in the same C source file, and the other one is Copyco

650) this.width=650; "src=" Http://s5.51cto.com/wyfs02/M02/77/75/wKioL1ZoIcOTDBBOAAECfBygMsY426.jpg "title=" QQ picture 20151209204034.jpg "alt=" Wkiol1zoicotdbboaaecfbygmsy426.jpg "/>Address description1,R1--R5 all want a loopback0=10.10.x.x2, R1--R3 E0 address =1.1.123.x3, the R2--r4 WAN interface is 1.1.234.X/244,r4-r5 WAN Interface =1.1.45.x/24Bridge connection1, configure the Ethernet address of the R1