atoi介绍
atoi即ASCII to integer，是把字符串转换成整型数的函数。
atoi看起来是一道很简单的题目，但是却考察了面试者写代码过程的各方面的能力————对于测试用例全面的考虑、良好编程的习惯、对代码漏洞和鲁棒性的保证。
对于技术面试而言，应聘者应该养成在写代码之前考虑所有可能的测试用例的习惯，保持思维逻辑的严谨。
atoi要求
Implement atoi to convert a string to an integer
- 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.
第一版代码
C++代码
enum Status {kValid = 0, kInvalid};
int globalStatus = kValid;
class StringHandling{
public:
int atoi(const char *str){
int number = 0;
globalStatus = kInvalid;//value is not correct
if (str != NULL) //检测空指针
{
bool negative = false;
if(*str != '\0')//检测空字符串
{
if(*str == '+')
str++;
else if(*str == '-')
{
negative = true;
str++;
}
while(*str != '\0')
{
if (*str >= '0' && *str <= '9')
{
number = number*10 + (*str - '0');
str++;
}
else
{
number = 0;
break;
}
}
if(*str == '\0')
{
globalStatus = kValid;//输入为有效
if (negative == true)//负数的情况
{
number = 0 - number;
}
}
}
}
return number;
}
};
测试用例
- 功能测试：输入的字符串表示正数、负数和0
- 边界值测试：最大的正整数、最小的负整数
- 特殊输入测试：输入字符串为NULL指针、输入字符串为空字符串、输入的字符串中又非数字字符
StringHandling sh;
cout << sh.atoi("") << "\t";
cout << "global status is: " << globalStatus << endl;
cout << sh.atoi(NULL) << "\t";
cout << "global status is: " << globalStatus << endl;
cout << sh.atoi("2368d") << "\t";
cout << "global status is: " << globalStatus << endl;
cout << sh.atoi("65433") << "\t";
cout << "global status is: " << globalStatus << endl;
cout << sh.atoi("-65433") << "\t";
cout << "global status is: " << globalStatus << endl;
cout << sh.atoi("--65433") << "\t";
cout << "global status is: " << globalStatus << endl;
显示结果
0 global status is: 1
0 global status is: 1
0 global status is: 1
65433 global status is: 0
-65433 global status is: 0
0 global status is: 1
存在的问题
- 首先，这里定义了全局变量globalStatus来标记遇到非法输入的情况，虽然返回的整数是0，但当globalStatus为Invalid（Invalid=1）时，说明输入的字符串是非法操作。
- 当输入时空字符串时，返回值是0，但globalStatus没有设置为Invalid
- 当输入只有一个正号或者负号，后面没有跟数字，也不会设置globalStatus
- 像最大正整数和最小负整数这样的边界条件没有考虑
参考代码
class Solution{
public:
int atoi(const char *str)
{
long number = 0;
globalStatus = kInvalid;
if(str != NULL && *str != '\0')//检查NULL指针或者空字符
{
bool negative = false;
if(*str == '+'){
str++;
}
else if(*str == '-'){
str++;
negative = true;
}
if (*str != '\0') //检查只有"+"或"-"的情况
{
number = StrToNumCore(str, negative);
}
}
return (int)number;
}
};
long StrToNumCore(const char *digit, bool negative)
{
long num = 0;
while(*digit != '\0')
{
if(*digit >= '0' && *digit <= '9')
{
int flag = negative ? -1 : 1;
num = num*10 + flag*(*digit - '0');
if((negative && num < (signed int)0x80000000) //检查最大正整数和最小负整数的边界
|| (!negative && num > 0x7fffffff))
{
num = 0;
break;
}
digit ++;
}
else
{
num = 0;
break;
}
}//while
if(*digit == '\0')
{
globalStatus = kValid;
}
return num;
}
