关于#C++#的类型转换问题，如何解决？

Question

我想对各个数进行类型转换并且10进制16进制输出，但是结果看起来不太美妙，出现了一些奇怪的数字，我想知道哪里出现了问题，并且在移位运算那里不能够输出，报错了，请问是什么原因？代码应该怎么改才能正确？devC++

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

int main()
{
    short a=-12345;
    cout<<"----------short -12345----------"<<endl;
    cout<<"10进制 "<<setbase(10)<<(int)a<<" "<<(unsigned short)a<<" "<<(unsigned int)a<<" "<<(float)a<<" "<<endl;
    cout<<"16进制 "<<setbase(16)<<(int)a<<" "<<(unsigned short)a<<" "<<(unsigned int)a<<" "<<(float)a<<" "<<endl;
    
    int b=2147483647;
    cout<<"----------int 2147483647----------"<<endl;
    cout<<"10进制 "<<setbase(10)<<(short)b<<" "<<(unsigned short)b<<" "<<(unsigned int)b<<" "<<(float)b<<" "<<endl;
    cout<<"16进制 "<<setbase(16)<<(short)b<<" "<<(unsigned short)b<<" "<<(unsigned int)b<<" "<<(float)b<<" "<<endl;

    float c=123456.789e5;
    cout<<"----------float 123456.789e5----------"<<endl;
    cout<<"10进制 "<<setbase(10)<<(double)c<<endl;
    cout<<"16进制 "<<setbase(16)<<(double)c<<endl;
    
    double d=123456.789e5;
    cout<<"----------double 123456.789e5----------"<<endl;
    cout<<"10进制 "<<setbase(10)<<(float)d<<endl;
    cout<<"16进制 "<<setbase(16)<<(float)d<<endl;
    
    cout<<"----------移位操作----------"<<endl;
    cout<<"左移："<<endl; 
    cout<<"10进制 "<<setbase(10)<<(a<<2)<<" "((unsigned short)a<<2)<<endl; 
    cout<<"16进制 "<<setbase(16)<<(a<<2)<<" "((unsigned short)a<<2)<<endl; 
    cout<<"右移："<<endl; 
    cout<<"10进制 "<<setbase(10)<<(a>>2)<<" "((unsigned short)a>>2)<<endl; 
    cout<<"16进制 "<<setbase(16)<<(a>>2)<<" "((unsigned short)a>>2)<<endl; 
    return 0;
    
} 

这是运行结果，我觉得在转化为unsigned int那一栏的结果有问题，看起来很奇怪，数据太大，为什么？

Answer 1

你把结果贴出来，把你觉得有问题的指出来

Answer 2

我贴上了，大佬，麻烦帮我看看

Answer 3

负数转换为unsigned int类型，肯定是很大的值，因为负数最高位是1，这是符号位。但无符号整型是没有符号位的，所以最高位是有效数字位，因此整型值很大。-12345转换为无符号整型的结果，就是32位的最大值减去12345，40多亿是正确的结果