c++ decltype和auto对比学习

#include <iostream>

using namespace std;

int main()
{

    int a=1;
    int b=2;
    int &ra = a;
    int *ptr = &a;
    /*
    *1.decltype对于括号和解引用操作符*，会解析为引用＆
    *int &ra = a;
    *decltype(ra) da; //int &
    *decltype(a) db;  //int
    *decltype((a)) dc;//int &,必须初始化，否则报错
    */
    //  decltype(ra) refa; //int &,error:reference must initialized
    decltype(a) testb; //int
    //  decltype((a)) refb; //int &,error:reference must initialized
    //  decltype(*ptr) refc;//int &,error:reference must initialized

    /*
    *２．decltype和auto对于const的处理对比
    *decltype 保留底层const和顶层const
    *auto只保留底层const
    *auto不得不保留底层const是因为从底层const常量转换为没有底层const的变量是错误的
    *const int *cpa = &i;
    *auto a = cpa; //a是const int *
    */
    //顶层const,decltype保留，auto忽略
    const int cona = 1;
    decltype(cona) td = 1; // const int
    auto hd = cona;   //int
    hd = 1; //ok
    //  td = 1; //error:const int read-only

    //底层const，decltype保留，auto保留
    const int *conb = &a;
    decltype(conb) te = &a;//const int *
    //  *te = 2; //error:const int * read-only locatin
    auto fe = conb; //const int *
    //  *fe = 3; //error:assignment of read-only location

    //底层和顶层const
    const int * const conc = &a;
    decltype(conc) tf = &a; //const int * const
    //tf = &b; //error:assignment of read-only variable
    //*tf = 3; //error:assignment of read-only location
    auto ft = conc; //const int *
    ft = &b; //ok
    //  *ft = 4;  //error:assignment of read-only location
    return 0;
}