# 2018-2019-1 20165230 "Fundamentals of Information Security system Design" Third week study summary

2018-2019-1 20165230 "Fundamentals of Information Security system design" The third week study summary of learning Contents

• The actual implementation of the memory system is to combine multiple hardware memory and operating system software.
• How valid addresses are calculatedImm(Eb,Ei,s) = Imm + R[Eb] + R[Ei]*s

• ISA: Instruction set architecture, which defines the state of the processor, the format of the instruction, and the effect of each instruction on the state.

• If we want to get 32 bits of code on a 64-bit machine, you can use the instructionsgcc -m32 -S xxx.c

• The corresponding representation of the C language data type in X86
  

• gcc -S xxx.c -o xxx.sGet assembly code, also can be disassembled objdump -d xxx ;

• Stack frame: A process activity record is a data structure used by the compiler to implement a procedure/function call. Logically, the stack frame is a function of the yellow essence: function parameters, functions of local variables, functions after the execution of the return to where and so on.

• In the stack frame, the stack grows to the low address direction. So, if we want to increase the stack space, we have to do is to reduce the value of the stack pointer instead of increasing. Each invocation of each function has its own stack frame, which maintains all the necessary information. Register EBP points to the point of the current stack frame (high address), register ESP points to the top of the current stack frame (low address).

• Run the GCC compiler to produce a compilation file CODE.S:
  gcc -Og -S code.c

• With the-C command option, GCC compiles and assembles the code to get CODE.O:
  gcc -c code.c

Problems in teaching materials learning and the solving process

• What is the difference between the question 1:leap directive and the MOVQ directive?
• Problem 1 Solution: the MOVQ directive belongs to the Mov class (the simplest form of data transfer instruction) and MOVQ represents 4 bytes passed. Leap is read from memory to register, but does not actually reference memory.
• Question 2: -0g What is the optimization level and what are the optimization levels?

• Problem 1 Solution:
  The compile option-og tells the compiler to use an optimization level that generates machine code that conforms to the overall structure of the original C code. Higher level optimizations (-O1 or-O2) from the resulting program performance considerations

  Problems in code debugging and the resolution process

  Not currently

Code Hosting

Learning progress Bar

	lines of code (new/cumulative)	Blog Volume (Add/accumulate)	Learning Time (new/cumulative)	Important Growth
Goal	5000 rows	30 Articles	400 hours
Second week	30	1/1	8
Third week	198/228	2/3	12/20
Week Four
Week Five

Reflection

The recent learning situation is very bad, the attitude towards learning has also been lax. Last week, I sounded a wake-up call, in the following days, to take good care to make up for the debts owed.

# 2018-2019-1 20165230 "Fundamentals of Information Security system Design" Third week study summary