写一个函数，检查字符是否是整数，如果是，返回其整数值（或：怎样只用4行代码编写出一个从字符串到长整型的函数）

方法1：

#include <stdlib.h>#include <stdio.h>#include <string>#include <iostream>using namespace std;long strtoint(char *str,int length);int main(int argc, char* argv[]){int i=0;char str[100];while ((str[i]=getchar())!='0') {i++;}long n=strtoint(str,i);cout<<n<<endl;return 0;}long strtoint(char *str,int length){if(length > 1) {return str[0]=='-' ? strtoint(str, length-1)*10-(str[length-1]-'0') : strtoint(str, length-1)*10+str[length-1]-'0';} else {return str[0]=='-' ? -1/10 : str[0]-'0';}}

方法2：

#include <stdlib.h>#include <stdio.h>#include <string>#include <iostream>using namespace std;int CovertStringToLon(char *str,long &lLongValue);int main(int argc, char* argv[]){char *str = "-123";long m;CovertStringToLon(str,m);cout<<"2  "<<m;system("pause");return 0;}int CovertStringToLon(char *str,long &lLongValue) { char *p = ((*str == '-'|| *str == '+') ? str+1:str);for(lLongValue = 0; *p >= '0' && *p <= '9' && *p != 0; p++) lLongValue = lLongValue * 10 + (*p- '0'); lLongValue = (*str== '-')?(-lLongValue):(lLongValue); return (*p == 0 ? 1:0); }