atoi和itoa函数及负数转正数时溢出问题

首先贴出c函数库里的atoi函数, 其实是atol函数, 因为atoi调用了atol函数. 函数很简单,相信大家一看就懂.

isspace函数是判断传入字符是否为空白符, 空白符指空格, 水平制表, 垂直制表, 换页, 回车和换行符.

函数首先跳过空白字符, 之后判断正负号, 判断完正负号后判断字符是否为数字, 如果是则循环, 直到遇到非数字为止, 如果第一次循环就不是数字则直接返回 total ,此时 total为 0.

atoi函数:

/****long atol(char *nptr) - Convert string to long**Purpose:*       Converts ASCII string pointed to by nptr to binary.*       Overflow is not detected.**Entry:*       nptr = ptr to string to convert**Exit:*       return long int value of the string**Exceptions:*       None - overflow is not detected.********************************************************************************/long __cdecl atol(        const char *nptr        ){        int c;              /* current char */        long total;         /* current total */        int sign;           /* if '-', then negative, otherwise positive */        /* skip whitespace */        while ( isspace((int)(unsigned char)*nptr) )            ++nptr;        c = (int)(unsigned char)*nptr++;        sign = c;           /* save sign indication */        if (c == '-' || c == '+')            c = (int)(unsigned char)*nptr++;    /* skip sign */        total = 0;        while (isdigit(c)) {            total = 10 * total + (c - '0');     /* accumulate digit */            c = (int)(unsigned char)*nptr++;    /* get next char */        }        if (sign == '-')            return -total;        else            return total;   /* return result, negated if necessary */}/****int atoi(char *nptr) - Convert string to long**Purpose:*       Converts ASCII string pointed to by nptr to binary.*       Overflow is not detected.  Because of this, we can just use*       atol().**Entry:*       nptr = ptr to string to convert**Exit:*       return int value of the string**Exceptions:*       None - overflow is not detected.********************************************************************************/int __cdecl atoi(        const char *nptr        ){        return (int)atol(nptr);}

关于itoa函数, 我会贴出三个版本, 第一个是网上找的, 第二个是c函数库里的, 第三个我自己写的(利用<<c陷阱和缺陷>>里的一段代码改的).
版本1:

char * my_itoa( int num, char*str, int radix ){    const char table[] = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ" ;    char*ptr=str ;    bool negative=false ;    if(num==0)    {        //num=0        *ptr++='0' ;        *ptr='/0' ;        // don`t forget the end of the string is '/0'!!!!!!!!!        return str ;    }    if(num<0)    {        //if num is negative ,the add '-'and change num to positive        *ptr++='-' ;        num*=-1 ;        negative=true ;    }    while(num)    {        *ptr++=table[num%radix];        num/=radix ;    }    *ptr='/0' ;    //if num is negative ,the add '-'and change num to positive    // in the below, we have to converse the string    char*start=(negative?str+1:str);    //now start points the head of the string    ptr--;    //now prt points the end of the string    while(start<ptr)    {        char temp=*start ;        *start=*ptr ;        *ptr=temp ;        start++;        ptr--;    }    return str ;}

其实, 这个版本的itoa函数有个很明显的错误, 相信看过 <<c陷阱和缺陷>> 的朋友已经看出来了, 那就是溢出的问题, 代码中有这么一句话 num* = -1,这句话是在num为负数时将num转化为正数, 但问题是计算机存储整数用的是补码形式, 这就导致负数的表示范围比正数大一点. 我们以 int 占4字节为例, 其中负数最小可以为 -2^31 = -2147483648. 而正数最大为 2^31-1 = 2147483647. 所以如果不幸的是我们传入函数的num是负数,并且是最小的负数,那么程序就会出问题了. 比如说如果num = -2147483648, 那么num*=-1,之后num的值还是为-2147483648, 没有变化, 大家可以查看汇编imul之后的结果. 这就导致之后求得的索引是乱码. 所以最后输出的字符串也是乱码.  那如何修改程序呢, 其实很简单,我们再定义一个unsigned int num_neagtive; 将num*=-1; 这句话换成 num_negative = (unsigned int )(-num_negative); , 并且将之后的num换成num_negative就好了.

版本2:

static void __cdecl xtoa (        unsigned long val,        char *buf,        unsigned radix,        int is_neg        ){        char *p;                /* pointer to traverse string */        char *firstdig;         /* pointer to first digit */        char temp;              /* temp char */        unsigned digval;        /* value of digit */        p = buf;        if (is_neg) {            /* negative, so output '-' and negate */            *p++ = '-';            val = (unsigned long)(-(long)val);        }        firstdig = p;           /* save pointer to first digit */        do {            digval = (unsigned) (val % radix);            val /= radix;       /* get next digit */            /* convert to ascii and store */            if (digval > 9)                *p++ = (char) (digval - 10 + 'a');  /* a letter */            else                *p++ = (char) (digval + '0');       /* a digit */        } while (val > 0);        /* We now have the digit of the number in the buffer, but in reverse           order.  Thus we reverse them now. */        *p-- = '\0';            /* terminate string; p points to last digit */        do {            temp = *p;            *p = *firstdig;            *firstdig = temp;   /* swap *p and *firstdig */            --p;            ++firstdig;         /* advance to next two digits */        } while (firstdig < p); /* repeat until halfway */}/* Actual functions just call conversion helper with neg flag set correctly,   and return pointer to buffer. */char * __cdecl _itoa (        int val,        char *buf,        int radix        ){        if (radix == 10 && val < 0)            xtoa((unsigned long)val, buf, radix, 1);        else            xtoa((unsigned long)(unsigned int)val, buf, radix, 0);        return buf;}

版本3:

char * itoa_modified( int    val,  char  *buf,  int    radix ){char *p, *firstdig;char temp;              /* temp char */p = buf;if ( val < 0 ) {*p++ = '-';}if ( val > 0 ) {val = -val;}firstdig = p;do {*p++ = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"[-( val % radix )];val = val / radix;} while ( val );*p-- = '\0';do {temp = *p;*p = *firstdig;*firstdig = temp; /* swap *p and *firstdig */--p;++firstdig; /* advance to next two digits */} while ( firstdig < p ); /* repeat until halfway */return ( buf );}

这里如果val是正数则将它转换为负数, 之后统一按负数处理. 这样就不会出现溢出问题了.