位域的规律题

有无unsigned、a:1 a:2 a:4 a:8 a:16 a:32 以及用char的.各值来试验就可发现位域的规律。

#include <iostream>

#include <windows.h>

using namespace std;

void main()

{

struct test

{

//在这里测试就可以发现其规律，有无unsigned、a:1 a:2 a:4 a:8 a:16 a:32 以及用char的各值来试验就可发现位域的规律

unsigned int a:8;

};

test t1;

t1.a = 0;

cout<<"十进制 -> 0: "<<dec<<t1.a<<endl;

cout<<"十六进制-> 0: "<<hex<<t1.a<<endl;

cout<<endl;

t1.a = 1;

cout<<"十进制 -> 1: "<<dec<<t1.a<<endl;

cout<<"十六进制-> 1: "<<hex<<t1.a<<endl;

cout<<endl;

t1.a = 2;

cout<<"十进制 -> 2: "<<dec<<t1.a<<endl;

cout<<"十六进制-> 2: "<<hex<<t1.a<<endl;

cout<<endl;

t1.a = 3;

cout<<"十进制 -> 3: "<<dec<<t1.a<<endl;

cout<<"十六进制-> 3: "<<hex<<t1.a<<endl;

cout<<endl;

t1.a = 126;

cout<<"十进制 -> 126: "<<dec<<t1.a<<endl;

cout<<"十六进制-> 126: "<<hex<<t1.a<<endl;

cout<<endl;

t1.a = 127;

cout<<"十进制 -> 127: "<<dec<<t1.a<<endl;

cout<<"十六进制-> 127: "<<hex<<t1.a<<endl;

cout<<endl;

t1.a = 128;

cout<<"十进制 -> 128: "<<dec<<t1.a<<endl;

cout<<"十六进制-> 128: "<<hex<<t1.a<<endl;

cout<<endl;

t1.a = 129;

cout<<"十进制 -> 129: "<<dec<<t1.a<<endl;

cout<<"十六进制-> 129: "<<hex<<t1.a<<endl;

cout<<endl;

t1.a = 253;

cout<<"十进制 -> 253: "<<dec<<t1.a<<endl;

cout<<"十六进制-> 253: "<<hex<<t1.a<<endl;

cout<<endl;

t1.a = 254;

cout<<"十进制 -> 254: "<<dec<<t1.a<<endl;

cout<<"十六进制-> 254: "<<hex<<t1.a<<endl;

cout<<endl;

t1.a = 255;

cout<<"十进制 -> 255: "<<dec<<t1.a<<endl;

cout<<"十六进制-> 255: "<<hex<<t1.a<<endl;

cout<<endl;

t1.a = 256;

cout<<"十进制 -> 256: "<<dec<<t1.a<<endl;

cout<<"十六进制-> 256: "<<hex<<t1.a<<endl;

cout<<endl;

t1.a = 257;