算法16：实现itoa和atoi

atoi 代码实现

//str:源字符串
//return:返回整数
int atoi(const char* str)
{
    assert(str != NULL);

    //符号：0为正数，1为负数
    int sign = 0;
    int result = 0;

    //去掉空格，制表符和换行符
    while(*str == ' '|| *str == '\t' || *str == '\n')
    {
        ++str;
    }
    //判断正负数
    if(*str == '-')
    {
        sign = 1;
        ++str;
    }
    else if(*str == '+')
    {
        ++str;
    }

    while(*str >= '0' && *str <= '9')
    {
        result = 10*result + *str - '0';
        ++str;
    }

    if(sign == 1)
    {
        result = 0 - result;
    }

    return result;
}

itoa代码实现

//num: 欲转换的数字
//dst: 转换后的字符串
//radix:转换后的进制数
char* itoa(int num,char* dst,int radix)
{
    //指针指向目标字符串
    char* p = dst;
    //无符号数字
    unsigned digval;
    //调转字符串参数
    char* firstDig = dst;
    //调转字符串临时便令
    char temp;

    if(num < 0)
    {
        *p++ = '-';
        firstDig++;
        num = 0 - num;
    }

    while(num > 0)
    {
        digval = num%radix;
        num = num/radix;

        if(digval > 9)
        {
            *p++ = (char)(digval - 10 + 'a');
        }
        else
        {
            *p++ = (char)digval + '0';
        }
    }

    *p-- = '\0';

    //倒转字符串
    while(firstDig < p)
    {
        temp = *firstDig;
        *firstDig = *p;
        *p = temp;

        firstDig++;
        p--;
    }

    return dst;
}

测试代码

int _tmain(int argc, _TCHAR* argv[])
{
    char str[10] = "-123434";
    int num = atoi(str);

    cout<<"-123434 num is "<<num<<endl;

    char dstStr[50] = {0};
    cout<<"-123434 radix = 10 :"<<itoa(num,dstStr,10)<<endl;
    cout<<"-123434 radix = 8 :"<<itoa(num,dstStr,8)<<endl;
    cout<<"-123434 radix = 16 :"<<itoa(num,dstStr,16)<<endl;
    cout<<"-123434 radix = 2 :"<<itoa(num,dstStr,2)<<endl;

    return 0;
}