我必须编写一个程序，将无符号十六进制转换为十进制，将十进制转换为十六进制，将有符号十六进制转换为十进制，将十进制转换为十六进制，而不使用流过滤器“十六进制”和“十进制”。我的程序适用于 hex 到 dec 和 dec 到 hex 的无符号转换，但不适用于 SIGNED hex 到 dec 和 dec 到 hex。任何帮助将不胜感激。提前谢谢您!
#include <iostream>
#include <string>
#include <cmath>
using namespace std;
void convertDecToHex(unsigned hexNumber);
int convertHexToDec(char hex[], unsigned numberSize);
signed SignedDecToHex (signed DecNumberSigned);
signed SignedHexToDec (int number);
int main (){
unsigned choice, numberSize = 0, hexNumber;
signed DecNumberSigned;
char hex[100000];
string number, HexNumberSigned;
while (1){
cout << "Enter 1 to convert from unsigned Dec to Hex:" << endl << "Enter 2 to convert from unsigned Hex to Dec:"
<< endl << "Enter 3 to convert from signed Dec to Hex:"<< endl << "Enter 4 to convert from signed Hex to Dec:" << endl ;
cin >> choice;
if (choice == 1){
cout << "Enter unsigned dec # to convert to hex: " << endl;
cin >> hexNumber;
convertDecToHex( hexNumber);
}//1
if (choice == 2){
cout << "Enter unsigned hex # to convert to dec: " << endl;
cin >> number;
numberSize = number.size();
strcpy(hex, number.c_str());
cout << convertHexToDec(hex, numberSize) << endl;
}//2
if (choice == 3){
cout << "Enter signed dec to convert to hex: " << endl;
cin >> DecNumberSigned;
convertDecToHex( SignedDecToHex(DecNumberSigned));
}//3
if (choice == 4){
cout << "Enter signed hex to convert to dec: " << endl;
cin >> HexNumberSigned;
numberSize = HexNumberSigned.size();
strcpy(hex, HexNumberSigned.c_str());
cout << SignedHexToDec(convertHexToDec( hex, numberSize)) << endl;
}//4
}//while
system ("pause");
}//main
void convertDecToHex(unsigned hexNumber){
int remainder, i=0;
unsigned m[99999], total, x;
char n [99999];
i=0;
do {
remainder = hexNumber % 16;
hexNumber /= 16;
m[i] = remainder;
i++;
}//doLooop
while (hexNumber > 0);
total = i;
for (i=0, x = total - 1; i < total; i++, x--){
if (m[i] == 1) n[x] = '1';
if (m[i] == 2) n[x] = '2';
if (m[i] == 3) n[x] = '3';
if (m[i] == 4) n[x] = '4';
if (m[i] == 5) n[x] = '5';
if (m[i] == 6) n[x] = '6';
if (m[i] == 7) n[x] = '7';
if (m[i] == 8) n[x] = '8';
if (m[i] == 9) n[x] = '9';
if (m[i] == 10) n[x] = 'A';
if (m[i] == 11) n[x] = 'B';
if (m[i] == 12) n[x] = 'C';
if (m[i] == 13) n[x] = 'D';
if (m[i] == 14) n[x] = 'E';
if (m[i] == 15) n[x] = 'F';
}
for (i=0; i < total; i++)
cout << n[i];
cout << endl;
}//DecToHex
int convertHexToDec(char hex[], unsigned numberSize){
int i, j;
int n[10000];
unsigned sum = 0;
char a,b,c,d,e,f;
for (i = 0 ; i < numberSize; i++){
if (hex[i] == '1') n[i] = 1;
if (hex[i] == '2') n[i] = 2;
if (hex[i] == '3') n[i] = 3;
if (hex[i] == '4') n[i] = 4;
if (hex[i] == '5') n[i] = 5;
if (hex[i] == '6') n[i] = 6;
if (hex[i] == '7') n[i] = 7;
if (hex[i] == '8') n[i] = 8;
if (hex[i] == '9') n[i] = 9;
if (hex[i] == 'a' || hex[i] == 'A') n[i] = 10;
if (hex[i] == 'b' || hex[i] == 'B') n[i] = 11;
if (hex[i] == 'c' || hex[i] == 'C') n[i] = 12;
if (hex[i] == 'd' || hex[i] == 'D') n[i] = 13;
if (hex[i] == 'e' || hex[i] == 'E') n[i] = 14;
if (hex[i] == 'f' || hex[i] == 'F') n[i] = 15;
}//forLoop
for (i = 0, j = numberSize - 1; i < numberSize , j >= 0 ; i++, j--)
sum += pow(16,i) * n[j];
return sum;
}//HexToDec
signed SignedDecToHex (signed DecNumberSigned){
signed convertNumber;
convertNumber = DecNumberSigned + (pow (2,16));
return convertNumber;
}// SignedDecToHex
signed SignedHexToDec (int number){
int newNumber;
newNumber = number - (pow (2,16));
return newNumber;
}//SingedHexToDec
最佳答案
不要使用 pow()在做整数运算时。 pow返回值为 double这是不准确的，并且在将其转换为 int 时可能导致值相差 1 .
理论上，当数字是 2 的幂时，双倍可能是准确的。但这假设您知道自己在做什么，而目前您还不知道 :-)
尽可能使用位移:
pow(2,n) --> 1 << n
如何实现pow(16, n)使用位移位作为练习留给读者。
提示:((i^n)^m) = i ^ (n*m)
关于c++ - 将有符号和无符号十六进制转换为 dec 和 dec 为十六进制 C++，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/19018952/