#include<iostream>#include<ctype.h>#include<math.h>#include<string>using namespace std;#define LETTER 0#define DIGIT 1#define POINT 2#define OTHER 3#define POWER 4#define PLUS 5#define MINUS 6#define EndState -1//12.34567e-123int w;//1234567int p;//123int n;//5int e;//-1int d;//记录数字double result;//记录最终结果char* all[] = {"12","34.567","89.",".345",".","12E34","12e34","12E+34","12e+34","12e-34","12.3E4","12.3e4","12.3E+4","12.3e+4","12.3E-4","12.3e-4",".38E4",".3e45",".38E+4",".3e+45",".38E-4",".3e-45","3.E45","38.e4","3.E+45","38.e+4","3.E-45","38.e-4"};int isNumber(char *w);//判断字符串w是不是无符号数int getChar(int c);//判断当前字符是数字,点,Ee,+,-void excute(int& CurrentState,int ch);//根据当前状态和当前字符类型来进入下一个状态int main(){int i = -1;for(;++i<sizeof(all)/sizeof(char*);){cout<<all[i]<<" : ";if(isNumber(all[i])){cout<<"YES ";cout<<result;}else{cout<<"NO";}cout<<endl;}return 0;}//判断字符串w是不是无符号数int isNumber(char *w){string s(w);int i = -1;int ch;int currentState = 0;::w = 0;p = 0;//123n = 0;//5e = 1;//-1result = 0.0;for(;++i<s.size();){ch = getChar(s[i]);excute(currentState,ch);if(currentState == EndState){return 0;}}switch(currentState){case 1:case 2:case 6:result = ::w * pow(10,e*p-n);return 1;default:return 0;};}/** * 1. c是数字,返回1 * 2. c是点，返回2 * 3. c是E，返回4 * 4. c是+号，返回5 * 5. c是-号，返回6 *///判断当前字符是数字,点,Ee,+,-int getChar(int c){switch(c){case '.':return POINT;case 'E':case 'e':return POWER;case '+':return PLUS;case '-':return MINUS;default:if(isdigit(c)){d = c - 48;return DIGIT;}break;}return OTHER;}/** * 状态有 * -1: EndState * 0 : * 读取 数字 -> 1 * 读取 点 -> 3 * * 1 : * 读取 数字 -> 1 * 读取 '.' -> 2 * 读取 'E' -> 4 * 2 : * 读取 数字 -> 2 * 读取 E -> 4 * 3: * 读取 数字 -> 2 * 4: * 读取 +/- -> 5 * 读取 数字 -> 6 * 5: * 读取 数字 -> 6 * 6: * 读取 数字 -> 6 *///根据当前状态和当前字符类型来进入下一个状态void excute(int& currentState,int ch){switch(currentState){case 0:switch(ch){case DIGIT:w = w * 10 + d;currentState = 1;break;case POINT:currentState = 3;break;default:currentState = EndState;break;};break;case 1:switch(ch){case DIGIT:w = w * 10 + d;currentState = 1;break;case POINT:currentState = 2;break;case POWER:currentState = 4;break;default:currentState = EndState;break;};break;case 2:switch(ch){case DIGIT:w = w * 10 + d;++n;currentState = 2;break;case POWER:currentState = 4;break;default:currentState = EndState;break;};break;case 3:switch(ch){case DIGIT:w = w * 10 + d;++n;currentState = 2;break;default:currentState = EndState;break;};break;case 4:switch(ch){case DIGIT:p = p * 10 + d;currentState = 6;break;case PLUS:currentState = 5;break;case MINUS:e = -1;currentState = 5;break;default:currentState = EndState;break;};break;case 5:switch(ch){case DIGIT:p = p * 10 + d;currentState = 6;break;default:currentState = EndState;break;};break;case 6:switch(ch){case DIGIT:p = p * 10 + d;currentState = 6;break;default:currentState = EndState;break;};break;};}
第一列 待判定的数字字符串
第二列 是否为无符号数
第三列 输出double FCON = w * pow( 10 , e * p – n); 即将字符串转换为double