Execution of procedures

CPU When the command is executed ：

1、 First ,CPU Read program counter （PC,program counter） The instruction that the pointer points to , Import it into the instruction register （IR, Instruction Register）, say concretely , The completion of the read instruction has 3 A step ：

1. CPU The control unit operation address bus specifies the memory address to be accessed .（ Simple understanding , Is to put PC The value in the pointer is copied to the address bus ）.
2. CPU Notify the memory device to prepare data .
3. CPU After receiving the data from memory , Store this data in the instruction register .

Complete the above 3 Step ,CPU Read successfully PC The pointer points to the instruction , Stored in the instruction register .

However ,CPU Analyze the instructions in the instruction register , Determine the type and parameters of the instruction . If it's a compute type instruction , Then it's up to the logic unit to calculate ; If it's a storage type instruction , So it's executed by the control unit .

PC The pointer increases by itself , Ready to get the next instruction .

For example 32 On the machine , Instruction is 32 position 4 Bytes , need 4 Memory address storage , therefore PC The pointer will increase automatically 4.

  With a=11+15 For example .

We will have a Review Next question ： Programs written by programmers a=11+15 yes character string ,CPU Cannot execute string , Only instructions can be executed . So a special program is needed here —— compiler . The core competence of the compiler is translation , It translates one program into another program language .

here , We need a compiler to translate the program written by the programmer into CPU Know the instructions （ Instructions are considered a low-level language , We usually write high-level languages ）.

Let's elaborate on a=11+15 Implementation process of ：

1. The compiler analyzes , Find out 11 and 15 Is the data （ The code will indicate the data type ）, So when the compiled program starts , A special area will be opened in memory to store such constants , This area is dedicated to storing constants , Data segment , As shown in the figure below ：

11 Stored in the address 0x100;

15 Stored in the address 0x104;

  

2. Compiler will a=11+15 Convert to 4 Orders , After the program starts , These instructions are imported into an area dedicated to storing instructions , That is, the body segment , As shown in the figure above , this 4 Instructions are stored in 0x200-0x20c In the area of ：

 0x200 Positional load The command will address 0x100 Data in 11 Import register R0;

0x204 Positional load The command will address 0x104 Data in 15 Import register R1;

0x208 Positional add Instruction will register R0 and R1 Add the values in , Storage register R2;

0x20c Positional store Instruction will register R2 The values in are stored back in the data area 0x1108 Location .

3. When it comes to concrete implementation ,PC The pointer first points to 0x200 Location , Then execute this in turn 4 Orders .

Here are a few more questions to explain ： 

1. Variable a It's actually an address in memory ,a It's a mnemonic for programmers .
2. Why? 0x200 The instructions representing loading data into registers in are 0x8c000100, We will discuss in detail below .

3. I don't know whether careful students find , In the example above , Every time we operate 4 An address , That is to say 32 position , This is because we are using 32 The seat is wide CPU give an example . stay 32 The seat is wide CPU in , So are the instructions 32 Bit . But the data can be less than 32 position , For example, you can add two 8 Bytes of bits .

4. About the content of data segment and body segment , I will continue to explain in the process and thread section of module 4 .