实现itoa()和atoi()

atoi()：将字符串转换为整型值
实现代码：

enum Status{valid,unvalid};int status = valid;//全局变量标记输入是否非法long long _AtoI(const char* str, bool minus){long long num = 0;while (*str != '\0'){//输入在0-9之间if (*str >= '0'&&*str <= '9'){int flag = minus ? -1 : 1;//判断正负num = num * 10 + flag*(*str - '0');//若溢出，输出0if ((!minus&&num > 0x7FFFFFFF) || (minus&&num < (signed int)0x80000000)){num = 0;break;}str++;}else{//非法输入，返回0num = 0;break;}}//此时*str=='\0'可以说明输入时合法的if (*str == '\0'){status = valid;}return num;}int AtoI(const char* str){status = unvalid;long long num = 0;//判断字符串是否有+-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 0;}

这里需要注意的是，当输入0和非法输入时，都会返回0，那么如何分辨是哪种输入返回的0呢？所以代码中定义了一个全局变量status，如果是非法输入，就将status置为unvalid;若想知道是哪种输入，查看status就可以知道了。
itoa()：将整型值转换为字符串。
代码实现：

//num：要转换的整数，str要写入转换结果的目标字符串,radix:要转换的进制char *ItoA(int num, char* str, int radix){    char index[37] = "0123456789abcdefghijklmnopqstuvwxyz";//索引表    unsigned mid;//中间变量    int i = 0,j,k;    if (radix == 10 && num < 0)//十进制负数    {        mid = (unsigned)-num;        str[i++] = '-';    }    else        mid = (unsigned)num;    do    {        str[i++] = index[mid % (unsigned)radix];        mid /= radix;    } while (mid);    if (str[0] == '-')        k = 1;    else        k = 0;    char temp;    for (j = k; j <= (i - k - 1) / 2.0; j++)    {        temp = str[j];        str[j] = str[i - j - 1];        str[i - j - 1] = temp;    }    return str;}