Implementatoito convert a string to an integer.
Hint:Carefully consider allpossible input cases. If you want a challenge, please do not see below and askyourself what are the possible input cases.
Notes:It is intended for thisproblem to be specified vaguely (ie, no given input specs). You are responsibleto gather all the input requirements up front.
spoilersalert… click to show requirements for atoi.
Requirements foratoi:
The function first discards as many whitespace characters as necessaryuntil the first non-whitespace character is found. Then, starting from thischaracter, takes an optional initial plus or minus sign followed by as manynumerical digits as possible, and interprets them as a numerical value.
The string can contain additional characters after those that form theintegral number, which are ignored and have no effect on the behavior of thisfunction.
If the first sequence of non-whitespace characters in str is not a validintegral number, or if no such sequence exists because either str is empty orit contains only whitespace characters, no conversion is performed.
If no valid conversion could be performed, a zero value is returned. Ifthe correct value is out of the range of representable values, INT_MAX(2147483647) or INT_MIN (-2147483648) is returned.
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#pragma once#include<iostream>using namespace std;int My_atoi(const char* pstr)//指针指向的内容不能变，指针不能变应是char* const pstr{if (!pstr){printf("Pointer is NULL\n");return 0;}long long result = 0;//用long long便于溢出处理，用unsigned int仍有可能溢出，因为2^32-1除以2^31-1 小于10int sign = 1;//标志符号，默认为正int out = false;//aoti()要求第一个数字或符号与最后一个数字或符号间只能是数字，当遇到第一个数字或符号后，out=true，再遇到非数字退出while (*pstr){if (*pstr > '9' || *pstr < '0')//开始转换后遇到非数字就返回if (out)return result*sign;if (*pstr == '-'){sign = -1;out = true;}else if (*pstr == '+')out = true;else if (*pstr <= '9' && *pstr >= '0'){result = result * 10 + *pstr – '0';//溢出时返回上下界if (result >= 2147483647 && sign == 1)return 2147483647;if (result >= 2147483648 && sign == -1)//若写成result*sign > 2147483648，，会自动将2147483648的二进制表示（无符号的），//与int比较就将该表示转化为有符号整数100000…即 -2147483648,所以此处应用unsigned int//若用int则2147483648变为-2147483648比较，其次int最大2147483647，就算不变也不对return -2147483647 – 1;//不能写成-2147483648out = true;}else if (*pstr != ' ' && *pstr != '\n' && *pstr != '\r'&&*pstr != '\t'&&*pstr != '\v'&&*pstr != '\f')//这些个与空格等价return result*sign;pstr++;}return result*sign;}void main(){char* s1 = "\t\f\v\n\r-00100\n\t\f\v\n\r1234";char* s2 = "–099";char* s3 = "s100";char* s4 = "+2147483647sc";char* s5 = "20.0sc";char* s6 = "20-1sc";char* s7 = "0-099";char* s8 = "-9999999999999999999";//溢出char* s9 = "+9999999999999999999";//溢出char* s10 = " 10522545459";//溢出int i1 = atoi(s1);int i2 = atoi(s2);int i3 = atoi(s3);int i4 = atoi(s4);int i5 = atoi(s5);int i6 = atoi(s6);int i7 = atoi(s7);int i8 = atoi(s8);int i9 = atoi(s9);int i10 = atoi(s10);cout << i1 << ' ' << i2 << ' ' << i3 << ' ' << i4 << ' '<< i5 << ' ' << i6 << ' ' << i7 << ' ' << i8 << ' ' << i9 << ' ' << i10 << endl;i1 = My_atoi(s1);i2 = My_atoi(s2);i3 = My_atoi(s3);i4 = My_atoi(s4);i5 = My_atoi(s5);i6 = My_atoi(s6);i7 = My_atoi(s7);i8 = My_atoi(s8);i9 = My_atoi(s9);i10 = My_atoi(s10);cout << i1 << ' ' << i2 << ' ' << i3 << ' ' << i4 << ' '<< i5 << ' ' << i6 << ' ' << i7 << ' ' << i8 << ' ' << i9 << ' ' << i10 << endl;int a = 2147483648;cout << a << endl;//-2147483648system("pause");}
午餐，晚餐。或许吃得不好，可是却依旧为对方擦去嘴角的油渍。
