Lesson 4 knowledge summary

define Define constants and macros

define Define constants

define Define constants , Constant here refers to literal constant , Any constant can be defined , Such as ：

#include <stdio.h>
#define MAX 25000
#define STR "Hello,CSDN"

Like the above two , Can be successfully compiled , Of course , There are other definitions , Not one by one .

define Defining macro

Such as ：

#include <stdio.h>
#define MAX(x,y) ((x)>(y)?(X):(Y));

It is equivalent to

MAX (int x,int y){
    
	int z=0;
	z=((x)>(y)?(x):(y));
	return z;
}

therefore define Defining macros is a little different from functions , This one has no return type , also "x","y" There is no description of the type of . Of course define There are many other ways to define macros , Here's just a brief look at .

The pointer

Speaking of pointers , Most people think this is a difficult thing to overcome , But actually , It's not as hard as rumors say , It's all rumors .
Pointer it is related to memory , Learn about memory , Pointers can be better understood .
It's in the computer （ With 32 Position as an example ）, A memory unit is a byte , Similarly, a memory unit has an address .（ Here is why a memory unit is not a bit ,kb,mb, But a simple explanation of a byte ）

For example, define a char a=‘b’, as everyone knows a The memory occupied is one byte （ The previous articles mentioned , what Int,long All the memory occupied by ）, And a byte is 8bit, So if a memory unit is one bit , Then a character variable needs 8 A bit （ Address ）, An integer variable requires 4 Bytes , That is to say 32 The bit （ Address ）, This pair of address numbers will waste a lot . alike , A memory unit is 1kb,1mb It's too big , So a memory unit is a byte .

There are two ways to define pointers

1. int* p=a, It represents a pointer that defines an integer variable p, What's in it is a The address of
2. int a=9;int* p;*p=&a; It means the same as the above
   ad locum * yes ：* Dereference operator , It means through p Find the address stored inside p Point to

The size of the pointer variable

stay 32 The result of running in the bit platform is 4, stay 64 The result of bit platform operation is 8

The reason for the difference

First map This is a 32 Platform , stay 32 Bit platform , An address is made up of 32 individual 2 Base numbers make up （8 individual 16 Base number ）, And one 2 Hexadecimal number is 1bit, So an address takes up 4 Bytes （32bit=4byte）
Also in 64 In the platform , An address is made up of 64 individual 2 Base numbers make up , So an address takes up 8 Bytes （64bit=8byte）
Here the simple understanding of pointer is over

Structure

1. Its use
   Upper figure
   The method used is the same as above , Just a little understanding , If you need to explain it in detail, I will talk about it after the following article .
2. Two output methods of structure
   The first one is :
   
   The second kind :
   The second is output by function , It's not what I think , So I'm not familiar with the second one , So I didn't elaborate
3. Two methods of pointing to the structure
   This is the function point

student1.name   

This is a common point

4. Enter structure data
   
   Just a quick explanation , Because it's an array name , So the top name,sex,hometown Didn't take the address ( Because the array name is the address ), And integer needs to take the address ( Self understanding )