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largest element in the size class.2) separation adaptation :3) Partner system : A special case of separation adaptation, where each size class is a power of 2. The basic idea is to assume a heap size of 2m words, we maintain a separate free block for each block size 2k, where 0≤k≤m. The request block size is rounded up to the nearest power of 2. Initially, there is only one free block with a size of 2m characters.Key Benefits: Quick Search and Quick
implements a finite integer operation compared to a real integer operation, with some special properties: because of overflow, the expression x*x can produce negative numbers. However, both unsigned and complement operations satisfy many of the other properties of integer operations, including binding laws, commutative laws, and distributive laws. This allows the compiler to do a lot of optimizations. We've seen several clever ways to use bit-level arithmetic and arithmetic operations: (1) us
group-linked cacheAs with the direct map cache, group index bits identify groups.2. Row matching and word selection in group-linked cacheMore complex than direct-mapped caches3. Row substitution when missing in group-linked cachefully-connected cacheA fully-connected cache consists of a group (e=c/b) that contains all the high-speed slowdown.Basic structure:1. Group selection in the fully-connected cacheThere's only one group, so it's simple.2. Line matching and word selection in fully-connecte
processor design, it is often necessary to compare a signal to a number of possible matching signals to detect if a certain instruction code being processed is part of a class of instruction code.Sequential implementation of Y86-64Organize the processing into stages1. For OPL (integer and logical Operations), RRMOVL (register-register transfer) and IRMOVL (immediate count-register transfer)2. For RMMOVL and Mrmov3. For PUSHL and POPL4. For jump, call and RETTake a fingerDecodingPerformVisitWrit
and the number of cylinders in a disk is generally the same relationship? Because in the exercise, they are replaced directly with each other.Problem two, the simple way to estimate disk access time is to take the seek time by 2, but in exercise 6.3, so that the results of 16ms and the actual result 10ms difference is very large?Harvest and experience:This time the homework is completed the longest time, spent three days, finally read the book carefully, the exercise. After finishing this, I ha
cause a set of states called deadlock zones.
Deadlock is a rather difficult problem because it is unpredictable.
Mutex lock Order rule : If for each pair of mutex (S,T) in the program, assign a whole order to all the locks, each thread requests the lock in this order, and is released according to reverse, the program is deadlock-free.How to resolve a deadlock
do not let a deadlock occur:
Static strategy: Design appropriate resource allocation algorithm, do not le
, can be divided into three cases: condition One: Normalized value (when the Order field is not all 0 or(3) Rounding of floating-point numbersThere are four scenarios: rounding to even numbers (default), rounding to 0, rounding down, rounding up.(4) floating point arithmeticFloating-point addition: does not satisfy the binding, satisfies the monotonic floating-point multiplication: does not satisfy the binding, satisfies the monotonicity, does not satisfy the distributive in the additionAll is 1
s=2^s cache groups, each containing an e cache line, each row consists of a b=2^b byte block, a valid bit (indicating whether the row contains valid information), t=m-(B+s) Tag bits, which uniquely identify blocks stored in this cache line
The cache determines whether a request is hit and then jerks the process of the requested word, divided into: Group selection, line matching, word extraction
Group selection: Extracts the group i
unsigned number to a larger data type, representing the beginning plus 0.Symbol extension-Converts a complement number to a larger data type.Three, integer arithmeticUnsigned number additionAn arithmetic operation overflow that indicates that a complete integer result cannot be placed into the word length limit of the data type.Multiply Constants either unsigned or complement operations, multiplying by a power of 2 results in overflow. Four, floating point numberWhen a number cannot be accurate
Information Security System Design Foundation Fourth Week study summaryLearning tasks: Textbook Chapter IIIDuration of study: 10 hoursLearning content
First, the textbook knowledge carding1. Program coding and machine-level code Program Code :gcc compiler, converting source code into executable code, c preprocessor-assembler-linker machine-Level code :The
2018-2019-1 20165336 "Information Security System Design Basics" week 5th Study Summary 1. learned knowledge points
SRAM and DRAM are both easy to lose, but they have different structures, different speeds, and faster access speed.
The prom can only be programmed once.
Programs stored in ROM devices are usually called firmware.
The EEPROM can era
20145321 "The basis of information security system Design" 7th Week study summary textbook study content summary sixth chapterA memory system is a hierarchical structure of storage devices with different capacities, costs, and access times.6.1 Storage TechnologyThree common storage technologies: ram/rom/disk(1) Random
20145235 "Fundamentals of Information Security system design" NO. 06 Week study summary _014.1.4 Y86 anomalySome values for the visible status Code stat:1:AOK program performs normally2:HLT indicates that the processor executed a halt instruction3:adr means that the processor reads from an illegal memory address or writes to an illegal memory address.4:ins indica
-20165206 2018-2019-1 Summary of the fifth week of information security system design-Summary of Teaching Materials-Random Access to memory:
Random Access Memory is divided into two types: static RAM (SRAM) and dynamic RAM (Dram ).
Static RAM stores each bit in a bistability memory unit, and dynamic RAM stores each bit to charge a capacitor.
Static Ram is mai
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 the block4. Deadlock1. A set of threads is blocked, waiting for a condition that will never be true.
Programmers use P and V to operate improperly so that the forbidden areas of two semaphores overlap.
Overlap
group
Send a signal using the/bin/kill program
Send a signal from the keyboard
Send a signal using the Kill function
Send a signal with the alarm function
(3) Receiving signalEach signal type has a pre-defined default behavior
Process termination
Process termination and dump memory
Process stops until the Sigcont signal is restarted
Process ignores this signal
(4) Signal processing problems
Pending signal is blocked
Pending signals are
are stored as contiguous sequence of bytes, with the address of the object being the smallest address in the byte being used.Linux 32, Windows, and Sun machines use 4-byte addresses, while Linux 64 uses 8-byte addresses.The ASCII code for the decimal digit x is exactly 0x3x, and the hexadecimal representation of the terminating byte is 0x00. Any system that uses ASCII code as a character code will get the same result, regardless of the byte order and
128 pages of reference books4. Translation Conditions BranchThe most common method of translating conditional expressions and statements from C to machine code combines conditional and unconditional jumps 5. Circulationthe assembly cycle can be combined with conditional tests and jumps to achieve6. Conditional Delivery InstructionsThe traditional way to implement conditional operations is to take advantage of controlled conditional shifts. Control transfer of actual use of datathe code that is
) or positive (s=0), and the sign bit with the value 0 is interpreted as a special case.Mantissa: M is a binary decimal order: E is weighted against floating-point numbers and can be negativeFloat:s=1 bit, exp=8 bit, frac=23 bitDouble:s=1 bit, exp=11 bit, frac=52 bit2. Integers and floating-point numbers indicate the relationship of the same number:P74: The relative region corresponds to the low of the integer, just before the highest significant bit equal to one stop, and a floating point repre
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