怎样将二进制补码转化为原码？
如，怎样将二进制补码1000001111000011010010010，0000101011110000101011110001010,101010111110101010001,01010100001111等化为原码的形式？
有现成的函数能直接转换吗？
[解决办法]
#include<climits>#include<string>#include<iostream>#include<bitset>using namespace std;#define BINARY_BIT (20)#define BINARY_SIGN (1<<(BINARY_BIT-1))#define BINARY_MASK (~-(1<<BINARY_BIT))string true_form_long(const string& complement){ bitset<sizeof(long)*8> bits(complement); unsigned long val = bits.to_ulong(); if(val > (unsigned long)(BINARY_SIGN-1)) { val=(val&(BINARY_SIGN-1)); return bitset<sizeof(long)*8>(((~val+1)|BINARY_SIGN)&BINARY_MASK).to_string().substr(32-BINARY_BIT); } if(complement.size()<BINARY_BIT) { string s(BINARY_BIT-complement.size(),'0'); return s+complement; } return complement;}string not_code(const string & complement){ bitset<sizeof(long)*8> bits(complement); unsigned long val = bits.to_ulong(); return bitset<sizeof(long)*8>((~val)&BINARY_MASK).to_string().substr(32-BINARY_BIT);}void test_long(long val){ cout<<"补码数字:"<<val<<"\n"; bitset<32> a(val); cout<<"补码:"<<a.to_string().substr(32-BINARY_BIT)<<"\n"; cout<<"原码:"<<true_form_long(a.to_string())<<"\n"; cout<<"反码:"<<not_code(a.to_string())<<endl; // bitset<32> b(true_form_long(a.to_string())); // cout<<b.to_string()<<endl;}void test_string(const string& str){ if(str.size()==32) cout<<"补码:"<<str.substr(32-BINARY_BIT)<<"\n"; else { string s(BINARY_BIT-str.size(),'0'); cout<<"补码:"<<s+str<<endl; } cout<<"原码:"<<true_form_long(str)<<"\n"; cout<<"反码:"<<not_code(str)<<endl;}int main(){ string str="1000001111000011010010010"; bitset<32> a(str); cout<<a.to_ulong()<<endl; cout<<true_form_long(str)<<endl; for(long i=-10;i<0;i++) { test_long(i); cout<<endl; } puts("输入补码"); string s; cin>>s; test_string(s);}