面试题33：把字符串转换为整数
C语言的库函数atoi()的作用是将一个字符串转换为整数。写一个函数StrToInt，实现这一功能。
// 写一个函数StrToInt实现将字符串转换为整数的功能.cpp : 定义控制台应用程序的入口点。#include "stdafx.h"#include <iostream>using namespace std;//特殊情况：1.字符串为空指针或空串//2.字符串中有非0到9的字符(负号,#，@，等等)//特别需要注意的是：还要考虑溢出的问题。正整数最大值0x7FFFFFFF,负整数最小值0x80000000//函数功能：将字符串str转换为整数, 由参数nInt接收整数值，若转换成功返回true,否则返回falsebool StrToInt(char *str, int &nInt){if (str == ""){cout << "空串！" << endl;return false;}else if (str == NULL){cout << "空指针！" << endl;return false;}else if ((strcmp(str, "+")==0) || (strcmp(str, "-")==0)){cout << "字符串输入无效！" << endl;return false;}else{char *p = str;bool isFirst = true;//标记是否为字符串的第一个字符bool hasMinus = false;//标记字符串的第一个字符是否为负号nInt = 0;while (*p != '\0'){if (isFirst && (*p)=='-')//有负号{hasMinus = true;p++;continue;}else if (isFirst && (*p)=='+'){p++;continue;}if ((*p) >='0' && (*p) <= '9')//当前字符为数字字符{nInt = nInt * 10 + (*p) - '0';if ((!hasMinus && nInt>0x7FFFFFFF) || (hasMinus && nInt<(signed int)0x80000000))//注意此处(signed int){cout << "字符串数值溢出，输入无效！" << endl;return false;}p++;}else//当前字符为非数字字符{cout << "字符串中包含有非数字的字符,不能转换为数字！" << endl;return false;} }if (hasMinus)//字符串有负号{nInt = (-1) * nInt;}return true; } }int _tmain(int argc, _TCHAR* argv[]){int nTest1 = 100000000000000;int nTest2 = -1000000000000000; if(nTest1 > 0x7FFFFFFF) { cout << "上溢！" << endl; } if (nTest2 < (signed int)0x80000000) { cout << "下溢！" << endl; }int nInt = 0;char *str = NULL;if (StrToInt("123", nInt)){cout << nInt << endl;}if (StrToInt("", nInt))//空串{cout << nInt << endl;}if (StrToInt(str, nInt))//空指针{cout << nInt << endl;}if (StrToInt("-123", nInt)){cout << nInt << endl;}if (StrToInt("+123", nInt)){cout << nInt << endl;}if (StrToInt("-12#3@", nInt)){cout << nInt << endl;}if (StrToInt("0", nInt)){cout << nInt << endl;}if (StrToInt("+", nInt)){cout << nInt << endl;}if (StrToInt("100000000000000000000000000000000000000", nInt)){cout << nInt << endl;}if (StrToInt("-100000000000000000000000000000000000000", nInt)){cout << nInt << endl;}system("pause");return 0;}运行结果：
说明：在判断上溢、下溢时出现错误！不知缘由，请大牛指正！