Week 1 - Computer Architecture & C
冯·诺伊曼系统
从应用场景的角度来说，曾经计算机分为两种：
- Application Specific Computer
- General Purpose Computer
Application Specific Computer的bottleneck在于，Programming required major re-wiring.
为了改进这一点，冯·诺伊曼发明了von Neumann Architecture system，是stored-program computer， a computer that stores program instructions in memory.
VNA的优点就是Re-prrogramming doesn't requrire any hardware modifications
CPU
由两个主要部件组成：Control Unit(CU) & Arithmetic and Logic Unit(ALU)
ALU负责实施具体的运算。
CU负责：
- Retrieves the operands
- Ask ALU to compute symbol "*" or "+" ..
- Store the result
- Jumps to next line
ALU
ALU是执行arthmetic和logical operations的电路的union。
CU
负责程序运行阶段指令一条一条的执行。
Register
Register是靠近ALU的small storage elements。
register用于暂存计算中的数据
ALU可以很快从register中读取数据
register的优点
VNA computer的三个主要部件中没有寄存器，可以用Memory去存数据。缺点就是内存的读写相比register的读写还是慢，且对于多次使用的元素，每次使用都要读取。用寄存器的话可以把多次要使用到的元素存在寄存器中，读取一次，而ALU从register中读data是0 time overhead的，所以更快。
改进后CPU
Memory
内存用于存both program instructions and data.
对于程序员来说，我们将memory视为storage element
Memory由很多cells组成，每个cell能存一小段数据，每个cell都有自己的address。 E.g. 0 ~ N-1
在数据read/write操作发生时，CPU生成要存/取的内存地址，Memory Management Unit(MMU，一个CPU模块)从对应的地址读写内容。
C语言中，用指针来操纵内存。
I/O interfaces
I/O用来接收和发送来自连接设备的信息。链接的设备（monitor, keyboard, mouse）叫作peripheral devices（外围设备）。
Run Program on VNA computer
main(){
readIO a;
readIO b;
c = a + b;
store c;
d = a - b;
print d;
}- 程序存在memory中
- CU读到readIO，理解到数据a需要由用户输入
- 用户从外围设备输入data，传到了ALU
- b也传到了ALU
- CU指挥ALU进行加运算，算出c
- CU将ALU中计算得到的c拷贝到Memory
- CU指挥ALU进行减操作，得到d
- CU将ALU中计算得到的d发送到外围设备display
C Program
由于作者比较熟悉这一块，很大部分内容都会被省略。具体内容请参考课堂slides
char c = 'A', d;这种写法中，d是un-initialized，包含garbage value initially。
Array size必须是constant，否则会causes malfunction
int array_size;
... // user input
int a[array_size] // 这样会报错
// ----------------
int array_size = 4; //这样不会报错二维数组在内存中的映射关系
字符串最后有\0
strcpy(dest, src) - copy a string strcat(str, "abc") - concatenate two strings strlen(str) - get string length strcmp(str1, str2) - compare two strings
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A byte consists of 2 Hex digits
a digit is called bit.
因为地址是8 bytes
栈用于存储函数的局部变量和上下文，它的操作是自动的；而堆用于存储程序运行时动态分配的内存，需要程序员手动管理。
char *ptr = "Comp Sc"; 在只读数据段，不能修改
char arr[] = "Comp Sc"; 在堆或者数据段，可以修改
Practice Problems
1.2. use the I/O interfaces to store a value in the registers.
- uses the ALU to compute the sum and stores the result in the register c.
- stores the value of the register c in main memory.
- uses the ALU to compute the difference and stores in the register d.
- uses the I/O interfaces to print the value of d.
Acces to registers inside the CPU is much faster than main mem- ory. You may save read and write-access to memory by using registers which is accessed frequently. The instructions given in the first ques-tion are an example: the registers a and b are used several times, saving memory access time.
CPU内部寄存器的访问速度比主存储器快得多。您可以通过使用频繁访问的寄存器来节省对内存的读写访问。
第一个问题中给出的指令是一个例子：寄存器a和b被多次使用，从而节省了内存访问时间。
Pointer store memory address.
Pointer and arrays can be transfered each other. The name of the array is a Pointer that point the address of the 1st unit. If you allocated an array by using pointer, you can also visit the i th unit by using *(p+i).
Conclusion: swap the value the pointer px and pointer py pointed to.
- What will be the output of this program?
int main () {
float arr[5] = {12.5,10.0,13.5,90.5,0.5};
float *ptr1 = &arr[0];
float *ptr2 = ptr1 + 3;
printf ("%f\n", *ptr2);
printf ("%ld\n", ptr2 - ptr1);
return 0;
}90.5
3
2, 1
String is stored as a continuous array of char in memory, 13bytes totle, 12 for hello world! and 1 for \0
Copy t string into s.
the same with 9
\
\
Larry Shi