如何将整数转换为二进制表示?
我正在使用下面的代码:
String input = "8";
String output = Convert.ToInt32(input, 2).ToString();
但是它抛出了一个异常:
找不到任何可解析的数字
如何将整数转换为二进制表示?
我正在使用下面的代码:
String input = "8";
String output = Convert.ToInt32(input, 2).ToString();
但是它抛出了一个异常:
找不到任何可解析的数字
当前回答
我在一个编码挑战中遇到了这个问题,你必须将32位十进制转换为二进制,并找到子字符串的可能组合。
using System;
using System.Collections.Generic;
using System.Globalization;
using System.Numerics;
using System.IO;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
namespace ConsoleApp2
{
class Program
{
public static void Main()
{
int numberofinputs = int.Parse(Console.ReadLine());
List<BigInteger> inputdecimal = new List<BigInteger>();
List<string> outputBinary = new List<string>();
for (int i = 0; i < numberofinputs; i++)
{
inputdecimal.Add(BigInteger.Parse(Console.ReadLine(), CultureInfo.InvariantCulture));
}
//processing begins
foreach (var n in inputdecimal)
{
string binary = (binaryconveter(n));
subString(binary, binary.Length);
}
foreach (var item in outputBinary)
{
Console.WriteLine(item);
}
string binaryconveter(BigInteger n)
{
int i;
StringBuilder output = new StringBuilder();
for (i = 0; n > 0; i++)
{
output = output.Append(n % 2);
n = n / 2;
}
return output.ToString();
}
void subString(string str, int n)
{
int zeroodds = 0;
int oneodds = 0;
for (int len = 1; len <= n; len++)
{
for (int i = 0; i <= n - len; i++)
{
int j = i + len - 1;
string substring = "";
for (int k = i; k <= j; k++)
{
substring = String.Concat(substring, str[k]);
}
var resultofstringanalysis = stringanalysis(substring);
if (resultofstringanalysis.Equals("both are odd"))
{
++zeroodds;
++oneodds;
}
else if (resultofstringanalysis.Equals("zeroes are odd"))
{
++zeroodds;
}
else if (resultofstringanalysis.Equals("ones are odd"))
{
++oneodds;
}
}
}
string outputtest = String.Concat(zeroodds.ToString(), ' ', oneodds.ToString());
outputBinary.Add(outputtest);
}
string stringanalysis(string str)
{
int n = str.Length;
int nofZeros = 0;
int nofOnes = 0;
for (int i = 0; i < n; i++)
{
if (str[i] == '0')
{
++nofZeros;
}
if (str[i] == '1')
{
++nofOnes;
}
}
if ((nofZeros != 0 && nofZeros % 2 != 0) && (nofOnes != 0 && nofOnes % 2 != 0))
{
return "both are odd";
}
else if (nofZeros != 0 && nofZeros % 2 != 0)
{
return "zeroes are odd";
}
else if (nofOnes != 0 && nofOnes % 2 != 0)
{
return "ones are odd";
}
else
{
return "nothing";
}
}
Console.ReadKey();
}
}
}
其他回答
using System;
class Program
{
static void Main(string[] args) {
try {
int i = (int) Convert.ToInt64(args[0]);
Console.WriteLine("\n{0} converted to Binary is {1}\n", i, ToBinary(i));
} catch(Exception e) {
Console.WriteLine("\n{0}\n", e.Message);
}
}
public static string ToBinary(Int64 Decimal) {
// Declare a few variables we're going to need
Int64 BinaryHolder;
char[] BinaryArray;
string BinaryResult = "";
while (Decimal > 0) {
BinaryHolder = Decimal % 2;
BinaryResult += BinaryHolder;
Decimal = Decimal / 2;
}
BinaryArray = BinaryResult.ToCharArray();
Array.Reverse(BinaryArray);
BinaryResult = new string(BinaryArray);
return BinaryResult;
}
}
转换。ToInt32(string, base)不会将基数转换为基数。它假设字符串包含一个以指定基数为底的有效数字,并转换为以10为基数。
所以你会得到一个错误,因为“8”不是一个以2为基数的有效数字。
String str = "1111";
String Ans = Convert.ToInt32(str, 2).ToString();
将显示15(1111以2为底= 15以10为底)
String str = "f000";
String Ans = Convert.ToInt32(str, 16).ToString();
将显示61440。
我在一个编码挑战中遇到了这个问题,你必须将32位十进制转换为二进制,并找到子字符串的可能组合。
using System;
using System.Collections.Generic;
using System.Globalization;
using System.Numerics;
using System.IO;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
namespace ConsoleApp2
{
class Program
{
public static void Main()
{
int numberofinputs = int.Parse(Console.ReadLine());
List<BigInteger> inputdecimal = new List<BigInteger>();
List<string> outputBinary = new List<string>();
for (int i = 0; i < numberofinputs; i++)
{
inputdecimal.Add(BigInteger.Parse(Console.ReadLine(), CultureInfo.InvariantCulture));
}
//processing begins
foreach (var n in inputdecimal)
{
string binary = (binaryconveter(n));
subString(binary, binary.Length);
}
foreach (var item in outputBinary)
{
Console.WriteLine(item);
}
string binaryconveter(BigInteger n)
{
int i;
StringBuilder output = new StringBuilder();
for (i = 0; n > 0; i++)
{
output = output.Append(n % 2);
n = n / 2;
}
return output.ToString();
}
void subString(string str, int n)
{
int zeroodds = 0;
int oneodds = 0;
for (int len = 1; len <= n; len++)
{
for (int i = 0; i <= n - len; i++)
{
int j = i + len - 1;
string substring = "";
for (int k = i; k <= j; k++)
{
substring = String.Concat(substring, str[k]);
}
var resultofstringanalysis = stringanalysis(substring);
if (resultofstringanalysis.Equals("both are odd"))
{
++zeroodds;
++oneodds;
}
else if (resultofstringanalysis.Equals("zeroes are odd"))
{
++zeroodds;
}
else if (resultofstringanalysis.Equals("ones are odd"))
{
++oneodds;
}
}
}
string outputtest = String.Concat(zeroodds.ToString(), ' ', oneodds.ToString());
outputBinary.Add(outputtest);
}
string stringanalysis(string str)
{
int n = str.Length;
int nofZeros = 0;
int nofOnes = 0;
for (int i = 0; i < n; i++)
{
if (str[i] == '0')
{
++nofZeros;
}
if (str[i] == '1')
{
++nofOnes;
}
}
if ((nofZeros != 0 && nofZeros % 2 != 0) && (nofOnes != 0 && nofOnes % 2 != 0))
{
return "both are odd";
}
else if (nofZeros != 0 && nofZeros % 2 != 0)
{
return "zeroes are odd";
}
else if (nofOnes != 0 && nofOnes % 2 != 0)
{
return "ones are odd";
}
else
{
return "nothing";
}
}
Console.ReadKey();
}
}
}
这是一个有趣的阅读,我正在寻找一个快速复制粘贴。 我知道我很久以前用bitmath做过这件事。
以下是我的看法。
// i had this as a extension method in a static class (this int inValue);
public static string ToBinaryString(int inValue)
{
string result = "";
for (int bitIndexToTest = 0; bitIndexToTest < 32; bitIndexToTest++)
result += ((inValue & (1 << (bitIndexToTest))) > 0) ? '1' : '0';
return result;
}
你可以在循环中加入一些模数。
// little bit of spacing
if (((bitIndexToTest + 1) % spaceEvery) == 0)
result += ' ';
你可能会使用或传入一个stringbuilder,并直接追加或索引,以避免释放,也可以绕过+=的使用;
int x=550;
string s=" ";
string y=" ";
while (x>0)
{
s += x%2;
x=x/2;
}
Console.WriteLine(Reverse(s));
}
public static string Reverse( string s )
{
char[] charArray = s.ToCharArray();
Array.Reverse( charArray );
return new string( charArray );
}