!!!! ANSWERED !!!!
我无需帮助完成简单的位操作程序。
我只需要对第5步提供一些帮助,因为您将从我的代码中看到我是一个完整的菜鸟,所以请不要笑:)任何帮助都将深受赞赏。
Console.WriteLine("Enter integer number");
int number = Convert.ToInt32(Console.ReadLine());
string binaryString = Convert.ToString(number, 2);
Console.WriteLine("The binary representation of {0} is", number);
Console.WriteLine(binaryString.PadLeft(16, '0'));
BitArray b = new BitArray(new int[] { number });
Console.WriteLine("Enter bit's position (0 to 15)");
int position = Convert.ToInt32(Console.ReadLine());
Console.WriteLine("Enter bit's value (true or false)");
bool value = Convert.ToBoolean(Console.ReadLine());
b.Set(position, value); //set value to given position based on input from the user
Console.WriteLine("Your changes transformed \n{0} \nto ",binaryString.PadLeft(16,'0'));
for (int i = 15; i >= 0; i--)
{
if (b[i] == true)
{
Console.Write(1);
}
else
{
Console.Write(0);
}
}
Console.WriteLine();
答案 0 :(得分:2)
Convert.ToInt32有一个重载,它接受用于整数
的字符串表示的基数 Console.WriteLine(Convert.ToInt32(binaryString, 2))
但是,在更改了BitArray中的位之后,要使用Convert.ToInt32,您需要将比特数组重新转换为字符串
此代码改编自another answer此处SO
.....
binaryString = ToBitString(b);
Console.WriteLine(Convert.ToInt32(binaryString, 2))
.....
public string ToBitString(BitArray bits)
{
var sb = new StringBuilder();
for (int i = bits.Count - 1; i>= 0; i--)
{
char c = bits[i] ? '1' : '0';
sb.Append(c);
}
return sb.ToString();
}
答案 1 :(得分:0)
将二进制表示存储到变量中:
string binary = binaryString.PadLeft(16, '0');
Console.WriteLine(binary);
然后得到位置和值:
Console.WriteLine("Enter bit's position (0 to 15)");
int position = Convert.ToInt32(Console.ReadLine());
Console.WriteLine("Enter bit's value (true or false)");
bool value = Convert.ToBoolean(Console.ReadLine());
进行更改:
var chars = binary.ToCharArray();
chars[position] = value ? '1' : '0';
binary = new string(chars);
显示新的二进制表示及其十进制等值:
Console.WriteLine("New binary value: {0}",binary);
Console.WriteLine("New decimal value: {0}", Convert.ToInt32(binary, 2));
答案 2 :(得分:0)
将最终循环更改为以下内容:
int v = 0;
for (int i = 15; i >= 0; i--)
{
if (b[i] == true)
{
Console.Write(1);
v += (int)Math.Pow(2, i);
}
else
{
Console.Write(0);
}
}