如何正确解释数字（hex，oct，dec） [英] How to interpret numbers correctly (hex, oct, dec)

问题描述

我试图编写一个程序，它接受 - 十六进制，八进制和小数 - 的输入，将它们存储在整数变量中，并将它们与转换为十进制形式一起输出。例如：

用户输入：0x43,0123,65

程序输出：

0x43十六进制转换为67十进制
0123八进制转换为83十进制
65十进制转换为65十进制

很明显，我需要一种解释数字的方法，但我不知道如何去做。我已经尝试了各种方法，如读取它们到一个函数，并将它们转换为字符串，反之亦然（参见这里代码示例），但是解释数字总是需要转换为一些格式，原始输入。

我唯一可以想到的是重载一个>>运算符，一次读取一个字符，如果它在输入的开头看到0x或0，将整个输入存储在字符串中，然后再读入int。然后程序会以某种方式在输出期间确定正确的操纵器。

不知道如果有一个更简单的方法来做到这一点，任何帮助是赞赏。

编辑：这已经解决了，但如果有人感兴趣，我决定发布代码。

 code> #includestd_lib_facilities.h
 
 void number_sys（string num，string& s）
 {
 if（num [0] =='0' &&&&（num [1]！='x'&&＃1]！='X'））s =octal 
 else if（num [0] =='0'&&（num [1] =='x'|| num [1] =='X'））s =十六进制 
 else s =decimal; 
} 
 
 int main（）
 {
 cout<< 输入十六进制，十进制或十进制数字。使用0xx取消。\\\
; 
 string a; 
 
 while（cin>> a）{
 if（a ==0xx）break; 
 string atype; 
 number_sys（a，atype）; 
 
 int anum = strtol（a.c_str（），NULL，0）; 
 
 cout<< a<< setw（20-a.length（））< atype<< setw（20）<< 转换为< setw（10）
< anum  setw（10）<< decimal\\\
; 
} 
 
 keep_window_open（）; 
} 
  

解决方案

  char * args [3] = {0x43，0123，65}; 
 for（int i = 0; i <3; ++ i）{
 long int value = strtol（args [i]，NULL，0） 
 printf（％s转换为％d decimal\\\
，args [i]，value）; 
} 
  

输出：

  0x43转换为67十进制
 0123转换为83十进制
 65转换为65十进制
  

I'm trying to write a program that takes input of - hexadecimals, octals, and decimals -, stores them in integer variables, and outputs them along with their conversion to decimal form. For example:

User inputs: 0x43, 0123, 65

Program outputs:

0x43 hexadecimal converts to 67 decimal
0123 octal converts to 83 decimal
65 decimal converts to 65 decimal

So obviously I need a way to interpret the numbers, but I'm not sure how to go about doing it. I've tried various methods such as reading them into a function and converting them into a string, and vice versa (see here for code examples), but interpreting the numbers always requires conversion to some format that trashes the original input.

The only thing I can think of is overloading a >> operator that reads a character at a time and if it sees 0x or 0 at the beginning of the input then it stores the whole input into a string before it is read into an int. Then the program would somehow have to determine the right manipulator during output.

Not sure if there is a simpler way to do this, any help is appreciated.

Edit: This has been solved, but I decided to post the code in if anyone is interested.

#include "std_lib_facilities.h"

void number_sys(string num, string& s)
{
  if(num[0] == '0' && (num[1] != 'x' && num[1] != 'X')) s = "octal";
  else if(num[0] == '0' && (num[1] == 'x' || num[1] == 'X')) s = "hexadecimal";
  else s = "decimal";
}

int main()
{
    cout << "Input numbers in hex, dec, or oct. Use 0xx to cancel.\n";
    string a;

    while(cin >> a){
    if(a == "0xx")break;
    string atype;
    number_sys(a, atype);

    int anum = strtol(a.c_str(), NULL, 0);

    cout << a << setw(20-a.length()) << atype << setw(20) << "converts to" << setw(10)
         << anum << setw(10) << "decimal\n";
                 }

    keep_window_open();
}

解决方案

Take a look at the strtol function.

char * args[3] = {"0x43", "0123", "65"};
for (int i = 0; i < 3; ++i) {
  long int value = strtol(args[i], NULL, 0);
  printf("%s converts to %d decimal\n", args[i], value);
}

Outputs:

0x43 converts to 67 decimal
0123 converts to 83 decimal
65 converts to 65 decimal

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