（自用）const修饰指针还是指向的变量-辨析

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#include<bits/stdc++.h>
using namespace std;

int main()
{
    int const a[]={1,2,3};
    
    cout<<a[0]<<" "<<a[1]<<" "<<a[2]<<"\n";//output:1 2 3
    //++a[0];//[Error] increment of read-only location 'a[0]'
    
    /*正确的，但不好复现 
    {
        const int * pp=a;
        cout<<pp<<" -> "<<*pp<<"\n";//output:0x73fde0 -> 1
        
        ++pp , cout<<pp<<" -> "<<*pp<<"\n";//output:0x73fde4 -> 2
        //int * mypp=pp-1;//[Error] invalid conversion from 'const int*' to 'int*' [-fpermissive]
        
        int * ppp = (int *)0x73fde0;
        cout<<ppp<<" -> "<<*ppp<<"\n";//output:0x73fde0 -> 1
        
        ++(*ppp);
        cout<<ppp<<" -> "<<*ppp<<"\n";//output:0x73fde0 -> 2
        cout<<a<<" -> "<<a[0]<<"\n";//output:0x73fde0 -> 2
        
        //可以看到，a[0]确实被改变了！！！ 
    }
    */
    
    
    {
        //int * const p=a;//[Error] invalid conversion from 'const int*' to 'int*' [-fpermissive]
        //无法声明！ 
        
        int b[]={100,200,300};
        int * const p=b;
        cout<<*p<<"\n";//output:100
        *p=1000 , cout<<*p<<" = "<<b[0]<<"\n";//output:1000 = 1000
        //++p;//[Error] increment of read-only variable 'p'
        
        //结论：p不可以变，但可以改变p指向的东西 
    }
    
    
    {
        int const *p=a;
        cout<<p<<" -> "<<*p<<"\n";//output:0x73fde0 -> 1
        //*p=100;//[Error] assignment of read-only location '* p'
        ++p , cout<<p<<" -> "<<*p<<"\n";//output:0x73fde4 -> 2
        //*p=100;//[Error] assignment of read-only location '* p'
        
        //结论：p可以随意变，但p指向的区域永远不能被修改（即：并不只是声明时指向的区域）
    }
    
    
    {
        int const * const p = &a[0];
        cout<<p<<" -> "<<*p<<"\n";//0x73fde0 -> 1
        //++p;//[Error] increment of read-only variable 'p'
        //++(*p);//[Error] increment of read-only location '*(const int*)p'
        
        //结论：p被const修饰，不可变；*p被const修饰，所以永远无法通过p改变p指向的区域 
    }
    return 0;
}