String to Integer (atoi)---题解

Description：
Implement atoi to convert a string to an integer.

Hint: Carefully consider all possible input cases. If you want a challenge, please do not see below and ask yourself what are the possible input cases.

Notes: It is intended for this problem to be specified vaguely (ie, no given input specs). You are responsible to gather all the input requirements up front.

Requirements for atoi:
The function first discards as many whitespace characters as necessary until the first non-whitespace character is found. Then, starting from this character, takes an optional initial plus or minus sign followed by as many numerical digits as possible, and interprets them as a numerical value.

The string can contain additional characters after those that form the integral number, which are ignored and have no effect on the behavior of this function.

If the first sequence of non-whitespace characters in str is not a valid integral number, or if no such sequence exists because either str is empty or it contains only whitespace characters, no conversion is performed.

If no valid conversion could be performed, a zero value is returned. If the correct value is out of the range of representable values, INT_MAX (2147483647) or INT_MIN (-2147483648) is returned.

https://leetcode.com/problems/string-to-integer-atoi/description/

总结：
1、学习了string中的这个函数：str.find_first_not_of(’ ‘);
2、注意有符号int的范围：
INT_MAX 2147483647
INT_MIN (long)-2147483648
计算中存储的二进制都是补码，因此32位有符号int的取值范围[-2147483648, 2147483647].
正数的反码和补码都与原码相同。
负数的反码为对该数的原码除符号位外各位取反。
负数的补码为对该数的原码除符号位外各位取反，然后在最后一位加1。
从补码求源码同原码求补码一致。
1000 0000，对应的原码为1 1000 0000，即-128.

更多参考：http://blog.csdn.net/diandianxiyu_geek/article/details/44098121

代码：

#include <iostream>#include <stdio.h>#include <stdlib.h>#include <string>#include <math.h>#include <algorithm>#define INT_MAX 2147483647#define INT_MIN (long)-2147483648using namespace std;int myAtoi(string str) {    int n=str.size(),sign=1;    long result=0;    int index=str.find_first_not_of(' ');    if(str[index]=='+'||str[index]=='-')        str[index++]=='+'?sign=1:sign=-1;    while('0'<=str[index]&&str[index]<='9'&&index<n)    {        result=result*10+(str[index++]-'0');        if(result*sign>INT_MAX) return INT_MAX;        if(result*sign<INT_MIN) return INT_MIN;    }    return result*sign;}int main(){    string s;    cin>>s;    cout<<myAtoi(s)<<endl;    return 0;}

这个代码参考：https://leetcode.com/problems/string-to-integer-atoi/discuss/

Question：这个代码在我的电脑上运行时，-2147483648前面不加long时，输入1，总输出-2147483648，不知道为什么???