如何判断全角半角

最新推荐文章于 2021-05-18 22:58:27 发布

原创最新推荐文章于 2021-05-18 22:58:27 发布 · 1.5k 阅读

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#null #delete #测试

本文介绍了一个简单的C++程序，用于统计字符串中的中文字符和英文字符数量。通过逐字符扫描并利用ASCII值判断字符类型的方法，实现了对输入字符串中不同类型的字符计数。

int IsChineseOrEnglish(LPCSTR pText)
{
unsigned char sqChar[10];
sqChar[0]=*pText;
sqChar[1]='/0';
if (sqChar[0]>=0xa1)
{
  if (sqChar[0]==0xa3)
  {
   return 1;//全角字符  }
  else
  {
   return 2;//汉字  }
}
else
{
  return 0;//英文,英文标点，数字

}
}
bool GetNumChineseAndNumEnglish(const char*pSrc, int &nNumChiRet, int &nNumEngRet)//获得汉字与英文的个数
{
if (pSrc == NULL)
{
  return false;
}
else
{
  int nLen = strlen(pSrc);
  char *pTxt = new char[nLen +1];
  char *pHeader = pTxt;
  //LPCSTR pTxt = pSrc;
  strcpy(pTxt, pSrc);
  pTxt[nLen] = '/0';
  while (*pTxt != '/0')
  {
   int nRet = IsChineseOrEnglish(pTxt);
   switch (nRet)
   {
   case 0:
    ++ nNumEngRet;
    pTxt++;
    break;
   case 1:
   case 2:
    ++ nNumChiRet;
    pTxt++;
    pTxt++;
    break;
   default:
    break;
   }

  }
  if (pHeader != NULL)
  {
   delete [] pHeader;
   pHeader = NULL;
  }
  return true;
}
}

下面是完整的测试程序：

#include <windows.h>
#include <iostream>
using namespace std;
int IsChineseOrEnglish(LPCSTR pText);
bool GetNumChineseAndNumEnglish(const char*pSrc, int &nNumChiRet, int &nNumEngRet);ÎÄµÄ×Ö·û¸öÊý
int main(int argc,char**argv)
{

int nNumChin = 0;
int nNumEng = 0;
GetNumChineseAndNumEnglish("£áÑîÐ¡Ó±Yanger7", nNumChin, nNumEng);
cout << "Number of Chinese:" << nNumChin << endl;
cout << "Number of English:" << nNumEng << endl;
return 0;
}

int IsChineseOrEnglish(LPCSTR pText)
{
unsigned char sqChar[10];
sqChar[0]=*pText;
sqChar[1]='/0';
if (sqChar[0]>=0xa1)
{
  if (sqChar[0]==0xa3)
  {
   return 1;  }
  else
  {
   return 2;//ºº×Ö
  }
}
else
{
  return 0; }
}
bool GetNumChineseAndNumEnglish(const char*pSrc, int &nNumChiRet, int &nNumEngRet){
if (pSrc == NULL)
{
  return false;
}
else
{
  int nLen = strlen(pSrc);
  char *pTxt = new char[nLen +1];
  char *pHeader = pTxt;
  //LPCSTR pTxt = pSrc;
  strcpy(pTxt, pSrc);
  pTxt[nLen] = '/0';
  while (*pTxt != '/0')
  {
   int nRet = IsChineseOrEnglish(pTxt);
   switch (nRet)
   {
   case 0:
    ++ nNumEngRet;
    pTxt++;
    break;
   case 1:
   case 2:
    ++ nNumChiRet;
    pTxt++;
    pTxt++;
    break;
   default:
    break;
   }

  }
  if (pHeader != NULL)
  {
   delete [] pHeader;
   pHeader = NULL;
  }
  return true;
}
}