Binary SHift

[Frank Stolfi]

I'm trying to figure out the left binary shift operator (line 40), but what I understand is when I shift it to the left the compiler automatically adds a 0 onto the end instead of rolling the first number over. I'm having some trouble understanding how to get the number to rollover instead of just adding a 0. Any help would be very much appreciated.

Danka,

  void mystery()
{
	ifstream inFile;
	char mysteryIn[5000];
	char keyEncrypt = 'X';
	int solution[5000];
	int solution2[5000];


	inFile.open("mystery.dat",ios::in | ios::binary );

	// Readin in File
	if(inFile.fail())
	{
			cout << "File did not open!";
			Sleep(2000);
			exit(1);
	}

	else
		{
			for(int i=0; i< 5000;i++)
				{
					inFile >> mysteryIn[i];
				}
		}
	//Decrypting Binary data
		for (int j = 0; j < 5000; j++)
			{
				solution[j] = mysteryIn[j] ^= keyEncrypt;			
			}
	//Reading out the first part of the mystery
			for (int m=0; m<5000; m++)
			{
				cout << (char)solution[m];
			}
	//Shiftiing 1 binary space to left and reading into solutions2 variable
		for (int l =0; l < 5000; l++)
		{
			solution2[l] = (solution[l] << 1);
		}	
	// Reading Solutions2 and typcasing it by character
		for (int k=0; k<5000; k++)
			{
				cout << (char)solution2[k] ;
			}
}

[giblit]

shifting a unit to the left one bit is the same as multiplying it by 2

Lets look at an example

4 << 1 == 8

4 in binary == 0100

When we shift it to the left one bit it is now 1000

Binary is base 2 so moving it one to the left is the same as multiplying it by 2

Basically this a << b == a * 2^b

*On a side note on line 40 you can also use the compound operator: <<= which would look like solution2[l] <<= 1;

[Frank Stolfi]

What about a binary shift left with wrap?

[giblit]

Actually I didnt notice that it is solution2 and solution so you can't use operator <<

Also where is the wrap? I don't see it.

Are you saying something like you want to have a 1 byte number ( 8 bits ) shift to the left and if its going to turn into a 9 bit to roll over to the first bit?

So say we have 128 == 1000 0000 and you try and shift it left one you want it to wrap to the first bit instead so make it 0000 0001?

You could just subtract 255 :P ( the 9th bit will be 256 and you are adding one to the first bit so 256 - 1 = 255 )

Say we have the number 223 == 1101 1111

Now you are trying to shift to the left and get 0001 1011 1110 == 446 but since it is now one bit over 8 bit you want it to wrap to the first bit

now it is 1011 1111 == 191

A way to check would be
446 - 255 == 191

Here's a quick demo:

	const int wrap = 255;

	int x = 1;
	
	std::cout << "X = " << x << std::endl;
	while( x < 256 ) 
	{
		x <<= 1;
		std::cout << "X = " << x << std::endl;
	}
	
	if( x >= 256 ) //the next shift
	{
		x -= wrap;
		std::cout << "X wrapped!" << std::endl << "X = " << x << std::endl;
	}

http://ideone.com/oEnRe4

[Frank Stolfi]

giblit - Thank you so much for the help. Yes. I want to wrap when I shift to the left, but I dont know how to code it. I understand your code, but how can I append it into my code?

Good God I feel like my brain is frying right now.

[giblit]

just put an if statement after line 40 within that for loop. If you're using a character they are 2 bytes. So you could use pretty similar to what I posted.

[Frank Stolfi]

so would would 'X' be solution[l]?

[giblit]

yes