模拟itoa

char* itoa(int num, char*str, int radix){/*索引表*/char index[] = "0123456789ABCDEF";unsigned unum;/*中间变量*/int i = 0, j, k;/*确定unum的值*/if (radix == 10 && num<0)/*十进制负数*/{unum = (unsigned)-num;str[i++] = '-';}else unum = (unsigned)num;/*其他情况*//*转换*/do{str[i++] = index[unum % (unsigned)radix];unum /= radix;} while (unum);str[i] = '\0';/*逆序*/if (str[0] == '-')k = 1;/*十进制负数*/else k = 0;char temp;/*for (j = k; j <= (i - 1) / 2; j++){temp = str[j];str[j] = str[i - 1 + k - j];str[i - 1 + k - j] = temp;}*/ j = i-1;while (k <= j){swap(str[k],str[j]);k++;j--;}return str;}int main(){char str[50];printf("%s", itoa(1233, str, 10));system("pause");}