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;Dir_ptr = Opendir (".");if (dir_ptr = = NULL) {Perror (".");Exit (1);}while ((DIRENTP = Readdir (dir_ptr)) = NULL)if (Direntp->d_ino = = Inode_to_find){strncpy (Namebuf, Direntp->d_name, Buflen);Namebuf[buflen-1] = ' + ';Closedir (DIR_PTR);Return}fprintf (stderr, "error looking for Inum%d\n", (int) inode_to_find);Exit (1);}ino_t Get_inode (char *fname){struct STAT info;if (stat (fname, info) = =-1) {fprintf (stderr, "cannot stat");Perror (fname);Exit (1);}return Info.st_ino;}Testioctl#include #

parameter is used in Linux and the general purpose of using the-L is to list the size of each file in blocks.The difference between the LS command and the Find command in 2.linuxThe Find command is to find the meaning of the lookup; When you know which file it is, you can find the sub-file with LS, but you can use the Find command to find the folder when you don't know what the drive letter is.Resources1. In-depth understanding of computer systems2.http://zhidao.baidu.com/question/588990569.htm

: Change of display tracking variable value Until jump out of the loop Finish Jump function Help Problems encountered 1. Static library? What is a static library?2. Where are the development tools and resources located? How to solve the Find related Blogs, write more detailedBuild and use static libraries and dynamic libraries in Linux under GCCHttp://blog.chinaunix.net/uid-235

register file.6) Update PC (PC Update, program counter): Sets the PC to the address of the next instruction.5, the instruction executes the procedure, takes out the instruction from the PC, then follows the above several steps to operate, repeats the cycle.6. The processor never needs to read the status updated by the instruction in order to complete the execution of an instruction.Resources: In-depth understanding of computer systems, reference sites:http://www.tuicool.com/articles/Zv6v6nInfor

entire wordSuppose we want to design a logic circuit to find the minimum value in a set of characters A,b,c, using the HCL to express is:int Min3 =[A b 1:C;];Two types of memory devices: Clock registers, random access memoryProcessing an instruction consists of a number of operations (six basic stages that can be organized into the calculations required to implement all Y86 directives):1. Take the finger2. Decoding3. Implementation4. Visit5. Write back6. Update your PCThe processor never needs

small-end method.Note: The destination address of the branch instruction and the calling instruction is an absolute address, rather than using a PC-relative addressing method as in IA32. The processor uses a PC-relative addressing method, and the branch instruction is encoded more succinctly, which also allows the code to be copied from one part of the memory to another without having to update all the branch destination addresses.For example:An important property of the ⑤ instruction set is th

where the contents of the disk sector should be stored. The logical block number translates a sector address and then transfers the content to main memory without CPU interference, a process known as direct memory access. This data transfer is called DMA transfer. When the volume of a disk sector is securely stored in main memory, the disk controller notifies the CPU by sending an interrupt signal to the CPU. Problem:may be a bit low-level, the book says that the main me

handler is called, the new signal screen word created by the system automatically includes the signal being delivered. This ensures that when a given signal is processed, if the signal occurs again, it will be blocked until the end of the processing of the previous signal; The response function is set to be valid and not resetThe implementation of sleep () should be divided into three steps: Register a signal signal (sigalrm,handler). Call the alarm () function. Pause () suspen

misses/number of references Hit Ratio = 1-No hit Hit time No hit penalty: Because of the extra time required to miss the 2. Specific impact: Cache Size: Hit ratio +, hit Time + Block Size: Spatial locality +, hit ratio +, number of cache rows-, time locality-, no hit penalty + Degree of coupling: E value big, jitter-, Price +, hit Time +, no hit penalty +, control logic + "compromise for not hit penalty low, low degree of coupling, not hit penalty high, use hig

that divides the cache into chunks, transfers blocks between different tiers, determines whether they are hit or miss, and processes themCache memoryGeneral-Purpose Cache Memory Architecture:Each memory address has a M-bit, which forms m=2^m different address cache groups: Array cache lines for S=2^s cache groups: b=2^b bytes of data blocks make up a valid bit: Indicates whether the row contains meaningful information marker bits: Uniquely identifies the block stored in this cache line, T=m-( B

status code stat from the result of the instruction execution, based on Icode,imem_error,instr_valid,dmem_error. update pc stage generates a new value for the program counter, depending on the type of instruction and whether to select the branch, the new PC may be valc, Valm, or Valp. Please explain why there is only one if statement in the C language code, and the assembly code contains two conditional branches?A: The first conditional branch is part of th

1. Machine-Level Code(1) Two kinds of abstract Defines the format and behavior of machine-level programs by ISA The memory address used by the machine-level program is the virtual address 2. Data format3. Operand designator4. Press in and eject stack data Follow the principle of first in and out Push Press in, pop delete Pushq press four words into the stack popq four words pop-up stack 5. Arithmetic and logical operations LEAQ Load Valid address INC plus a D

non-gate or non-gate HCL integer Expression Case Expression Format: [ select 1: expr 1 select 2: expr 2 . select k: expr k ] Set Relationship:iexp in{ iexp1,iexp2,...iexpk } Arithmetic/logic unit (ALU) Sequential implementation of Y86-64 Organize the processing into stages Value fetch--> decoding decode--> performing execute--> memory--> writeback write back write back--> update PC update SEQ

week's exam error summary 1.The following jump commands are related to ZF ()A. jmpB. JeC. jsD. JaE. JBF. JbeAnalytical:2.Assuming that the function of the C-expression T=a+b is completed with the add instruction, the correct statement about the condition Code Register is ()A. If t==0, then zf=1B. If tC. If tD. if (aE. if (aF. LEAQ directive does not affect the condition code registerG. CMP directives do not affect the condition code registerAnalysis: Textbook p135ZF: 0 logo. The result of the r

I. Learning Objectives Understanding the role of ISA abstraction Master Isa, and be able to learn other architecture extrapolate Understanding the pipeline and how it is implemented Second, the Learning content y86-64 directive MOVQ directive IRMOVQ rrmovq mrmovq RMMOVQ Four integer manipulation instructions Addq,subq,andq,xorq only the Register data 7 Jump Instructions Cmovle cmovl cmove cmovne cmovge CMOVG The call command returns the address to the stack, and then j

1.Y86-64 Instruction Set architecture①Y86-64 directive MOVQ directive IRMOVQ rrmovq mrmovq RMMOVQ Four integer manipulation instructions Addq,subq,andq,xorq only the Register data 7 Jump Instructions Cmovle cmovl cmove cmovne cmovge CMOVG The call command returns the address to the stack, and then jumps to the destination address, and the RET instruction returns from such calls Pushq and POPQ instructions are implemented into the stack and out of the stack Execution of Halt

operations until all of the request data is delivered. The UNIX kernel uses three related data structures to represent an open file. The table entries in the Descriptor table point to the table entries in the open file, and the table entries in the Open File table point to the table entries in the V-node table, each process has its own separate descriptor table, and all processes share the same open and V-node tables. Understanding the general composition of these structures allows us to c

work correctly for any viable trajectory. 2. Elimination Method:Dynamically assigns a separate block to each integer ID, and passes to the thread routine a pointer to the blockFive, Deadlock:A set of threads is blocked, waiting for a condition that will never be true.How to resolve a deadlockA. Do not allow deadlocks to occur:B. Let the deadlock occur:Resources:1. Teaching Materials: 11th chapter "Network Programming"2. Teaching Materials: 12th chapter "Concurrent Programming"20135223 He Weizin

multiplexing, and threading.Processes are automatically dispatched by the kernel, and because they have separate virtual address spaces, there must be an explicit IPC mechanism for sharing data.Regardless of the concurrency mechanism, synchronizing concurrent access to shared data is a difficult issue. The P and V operations on semaphores are proposed to help solve this problem.Concurrency also introduces some other difficult issues. A function called by a thread must have a property called thr

occurrence of competition:The correctness of a program relies on the x point of one thread to reach its control flow before another thread reaches the y point. That is, the programmer assumes that the thread will follow a particular trajectory through the execution state space, forgetting a guideline that the threaded program must work correctly for any viable trajectory. 2. Elimination Method:Dynamically assigns a separate block to each integer ID, and passes to the thread routine a pointer to