Question

我正在尝试编写一个输入 - 十六进制，八进制和小数 - 的程序，将它们存储在整数变量中，并将它们与它们转换为十进制形式一起输出。例如：

用户输入：0x43,0123,65

节目输出：

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

所以显然我需要一种解释数字的方法，但我不知道如何去做。我尝试了各种方法，例如将它们读入函数并将它们转换为字符串，反之亦然（请参阅here代码示例），但解释数字总是需要转换为某种格式来破坏原始输入。

我唯一能想到的是重载＆gt;＆gt;一次读取一个字符的运算符，如果它在输入的开头看到0x或0，那么它在将整个输入读入int之前将其存储到字符串中。然后程序将以某种方式在输出期间确定正确的操纵器。

不确定是否有更简单的方法可以做任何帮助。

编辑：这已经解决了，但如果有人有兴趣，我决定发布代码。

#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();
}

Answer 1

看一下strtol函数。

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);
}

输出：

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

Answer 2

您始终可以将其存储为要启动的字符串，并查看前两个字符以查看它们是否为0x：

std::string num;
std::cin >> num;

if (num[0] == '0' && num[1] == 'x')
{ 
  //handle
}

Answer 3

我不确定是否有C ++方法可以做到这一点，但是如果你不介意一点C-ishness，你可以将这个东西读成char数组并使用类似{{{ 1}}。 sscanf(buffer, "%i", &output)将输入解释为十六进制，八进制或十进制，具体取决于其格式，就像您描述的那样。

编辑：啊，不知道%i也可以这样做。不理我。

Answer 4

如果要保留基本信息（十六进制/八进制/十进制），则需要将该信息与整数值本身分开存储，并且需要至少解析输入的前几个字符string（sscanf（），strtol（）等不会为你保留这些信息）。

你可以滚动自己的迷你解析器来保存输入库并进行转换（代码偏离我的头顶，未经测试）：

char inputStr[MAX_INPUT_LENGTH+1];
char *p;
int result = 0;
char values[128];

/**
 * This enumeration serves double duty; it keeps track of what
 * base the input was entered in, and it controls the state machine 
 * used to parse the input; from a didactic POV, this is probably bad form
 */
enum {
  eStart, 
  eHexOrOctal, 
  eOctal, 
  eDecimal, 
  eHexadecimal, 
  eError
} eBase = eStart;


/**
 * Use the values array as a table to map character constants to their corresponding 
 * integer values.  This is safer than using an expression like *p - '0', in 
 * that it can work with character encodings where digits are not consecutive.
 * Yes, this wastes a little space, but the convenience makes
 * up for it IMO.  There are probably better ways to do this.
 */
values['0'] = 0; values['1'] = 1; values['2'] = 2; values['3'] = 3; 
values['4'] = 4; values['5'] = 5; values['6'] = 6; values['7'] = 7; 
values['8'] = 8; values['9'] = 9; values['a'] = 10; values['b'] = 11; 
values['c'] = 12; values['d'] = 13; values['e'] = 14; values['f'] = 15;

/**  
 * Insert code to get input string here 
 */

for (p = inputStr; *p != 0; p++)
{
  /**
   * Cycle through each character in the input string, adjusting the state
   * of the parser as necessary.  Parser starts in the eStart state.
   */
  switch(eBase)
  {
    /**
     * Start state -- we haven't parsed any characters yet
     */
    case eStart:
      if (*p == '0') eBase = eHexOrOctal; // leading 0 means either hex or octal
      else if (isdigit(*p))
      {
        eBase = eDecimal;    // leading non-0 digit means decimal
        result = values[*p];  
      }                    
      else eBase = eError;    // no other character may start an integer constant
      break;
    /**
     * HexOrOctal -- we've read a leading 0, which could start either a hex or
     * octal constant; we need to read the second character to make a determination
     */
    case eHexOrOctal:      
      if (tolower(*p) == 'x')  base = eHexadecimal;
      else if (isdigit(*p) && *p != '8' && *p != '9')
      {
        base = eOctal;   
        result = values[*p];
      }
      else eBase = eError;
      break;
    /**
     * Octal -- we are parsing an octal constant
     */
    case eOctal:
      if (isdigit(*p) && *p != '8' && *p != '9')
      {
        result *= 8;
        result += values[*p];
      }
      else eBase = eError;
      break;
    /**
     * Decimal -- we are parsing a decimal constant
     */
    case eDecimal:
      if (isdigit(*p))
      {
        result *= 10;
        result += values[*p];
      }
      else eBase = eError;
      break;
    /**
     * Hexadecimal -- we are parsing a hex constant
     */
    case eHexadecimal:
      if (isxdigit(*p))
      {
        result *= 16;
        result += values[tolower(*p)];
      }
      else eBase = eError;
      break;
    /**
     * String is not a properly formatted integer constant in 
     * any base; once we fall into the error state, we stay there.
     */
    case eError:
    default:
      break;
  }
}
if (eBase != eError)
{
  printf("input: %s ", inputStr); fflush(stdout);
  switch(eBase)
  {
    case eOctal: printf("octal "); break;
    case eHexadecimal: printf("hexadecimal "); break
    default: break;
  }
  fflush(stdout);
  printf("converts to %d decimal\n", result);
}
else
{
  /** Print a suitable error message here */
}

Answer 5

如果您确实使用单个整数变量来存储您显示最终输出所需的所有信息，那么您必须使用整数变量的一部分来存储输入所在的原始基础。否则它不可恢复。

如何正确解释数字（十六进制，八进制，十进制）

5 个答案: