String to Integer (atoi)

最新推荐文章于 2025-09-07 21:00:40 发布

转载最新推荐文章于 2025-09-07 21:00:40 发布 · 47 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/dylqt/p/4921520.html

文章标签：

#c/c++

本文详细解读了atoi函数的实现逻辑，包括如何跳过空格、处理符号、识别整数边界等关键点，并提供了可能导致错误处理的情况分析。

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.

Update (2015-02-10):
The signature of the C++ function had been updated. If you still see your function signature accepts a const char * argument, please click the reload button to reset your code definition.

spoilers alert... click to show requirements for atoi.

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.

主要是特殊情况的考虑

要跳过开头的空格
开头的“+”“-”表示为符号
在数字开头后中间有字母则停止，输出前面的数字
如果出现非法字符，全为空，或者没有内容则返回0
超过最大值，或者小于最小值；输出INT_MAX (2147483647) or INT_MIN (-2147483648)

int myAtoi(char* str) {
    long long result = 0;
    int flag = 1;
    while(*str == ' ')
        str++;
    if(*str == '+')
    {
        str++;
        if(*str > '9' || *str < '0')
            return 0;
    }
    else if(*str == '-')
    {
        str++;
        flag = -1;
        if(*str > '9' || *str < '0')
            return 0;
    }
    else if(*str > '9' || *str < '0')
        return 0;
    while(*str)
    {
        if(*str > '9' || *str < '0')
            break;
        result = result * 10 + *str - '0';
        str++;
        if(flag > 0 && result > 0x7fffffff)    //这里注意，放在里面是为了在数据过长就返回
            return 0x7fffffff;
        if(flag < 0 && result > 0x7fffffff)
            return 0x80000000;
    }
    
    return result * flag;
}