實作itoa()和atoi()

atoi()：将字元串轉換為整型值

實作代碼：

enum Status
{
valid,
unvalid
};
int status = valid;//全局變量标記輸入是否非法
long long _AtoI(const char* str, bool minus)
{
long long num = ;
while (*str != '\0')
{
//輸入在0-9之間
if (*str >= '0'&&*str <= '9')
{
int flag = minus ? - : ;//判斷正負
num = num *  + flag*(*str - '0');
//若溢出，輸出0
if ((!minus&&num > ) || (minus&&num < (signed int)))
{
num = ;
break;
}
str++;
}
else
{
//非法輸入，傳回0
num = ;
break;
}
}
//此時*str=='\0'可以說明輸入時合法的
if (*str == '\0')
{
status = valid;
}
return num;
}
int AtoI(const char* str)
{
status = unvalid;
long long num = ;
//判斷字元串是否有+-
if (str != NULL&&*str != '\0')
{
bool minus = false;
if (*str == '+')
str++;
else if (*str == '-')
{
str++;
minus = true;
}
if (*str != '\0')
num = _AtoI(str, minus);
}
return (int)num;
}

int main()
{
char* str1 = "12345";
char* str2 = "-12345";
char* str3 = "0";
char* str4 = "+";
char* str5 = "-";
char* str6 = "";
char* str7 = "999999999999999999999";
char* str8 = "-999999999999999999999";
char* str9 = "1a2bc3";

cout << str1 << "->" << AtoI(str1) << endl;
cout << str2 << "->" << AtoI(str2) << endl;
cout << str3 << "->" << AtoI(str3) << endl;
cout << str4 << "->" << AtoI(str4) << endl;
cout << str5 << "->" << AtoI(str5) << endl;
cout << str6 << "->" << AtoI(str6) << endl;
cout << str7 << "->" << AtoI(str7) << endl;
cout << str8 << "->" << AtoI(str8) << endl;
cout << str9 << "->" << AtoI(str9) << endl;
system("pause");
return ;
}
           

這裡需要注意的是，當輸入0和非法輸入時，都會傳回0，那麼如何分辨是哪種輸入傳回的0呢？是以代碼中定義了一個全局變量status，如果是非法輸入，就将status置為unvalid;若想知道是哪種輸入，檢視status就可以知道了。

itoa()：将整型值轉換為字元串。

代碼實作：

//num：要轉換的整數，str要寫入轉換結果的目标字元串,radix:要轉換的進制
char *ItoA(int num, char* str, int radix)
{
    char index[] = "0123456789abcdefghijklmnopqstuvwxyz";//索引表
    unsigned mid;//中間變量

    int i = ,j,k;
    if (radix ==  && num < )//十進制負數
    {
        mid = (unsigned)-num;
        str[i++] = '-';
    }
    else
        mid = (unsigned)num;
    do
    {
        str[i++] = index[mid % (unsigned)radix];
        mid /= radix;
    } while (mid);
    if (str[] == '-')
        k = ;
    else
        k = ;
    char temp;
    for (j = k; j <= (i - k - ) / ; j++)
    {
        temp = str[j];
        str[j] = str[i - j - ];
        str[i - j - ] = temp;
    }
    return str;
}