bits as an integer

Question

I have 31 bits (as a string) inside a textbox that I need to represent as an integer. My code is simple: myTextBox.Text = 1100111101000101000111010001011; myInt = int.Parse(myTextBox.Text); // error number too large Obviously I am going about this the wrong way. Any help would be appreciated.

Administrator · Answer

A little bit of integer conversation ... Twenty five years ago our teacher told us First ask: Why?. Numeric type uint will be the best number to store bit flags - it has no sign - and has 32 bits. Next - You use TextBox only for testing I think - such numer of signs is quite unreadable ...

Administrator · Answer

Ah my mistake the method actually takes a base 10 value which then AND'd later in code. So instead of passing 000010 I should be passing 2. :P Thanks anyway for the help. :)

Administrator · Answer

I need it for a COM object I am using - one method of the COM object requires I pass it an int which is AND'd with another value by the COM object. It's kind of like this: myComObject.setMask(int mask); where the mask is 31 bits (again which will be AND'd with another value later in code). So the way it is done is to pass up to 31 different 0's and 1's as an integer into the method. An example call would be myComObject.setMask(00001); // which IS different to the COM object than .setMask(1).

Administrator · Answer

Swamii The question about 3 posts back is the key. Changing the bits to an int in binary form will not preserve the individual bit values. You come up with a completely different number. The only way this would work assuming you could solve leading zero problems and what not would be to convert your binary int back to a string and parse out the bits Myself I think there is probably a better solution. I can't imagine why you would need an int in binary form in the first place. Bill F

Administrator · Answer

Int has no leading zero's ... Try to format text in TextBox using leading zero's: myTextBox.Text=myNumber.ToString(000000 ... 0000); ... myNumber=Convert.To...(myTextBox.Text); but the trouble with length of myNumber type persists. My next question: number is based 2 or not? If yes - try to convert number from 10 to 2 and vice versa: myTextBox.Text=(Convert.ToString(myNumber2)).PadLeft(myNumberLength'0'); ... myNumber=Convert.To...(myTextBox.Text2);

Administrator · Answer

Ok I let me change the wording of my problem. :) I can't preserve the leading 0's after converting several bits to an integer. Example: myStringOfBits = 0001; myInt = int.Parse(myStringOfBits); MessageBox.Show(myInt.ToString()); // output is 1 when I want it to be 0001

Administrator · Answer

My English isn't too good. Of course Int64 is about 19 long and is too short for Your first int which is about 29 long and for the second; 14 long it will be good. The longest types in .net as decimal or uint will be too short as well. May be the best solution is to come back to Convert.ToInt ... (compress) and use based 10 integers with converting them to strings based 2 (decompress) with: Convert.ToString(int2) or using strings without conversion.

Administrator · Answer

Int64.Parse didn't work; got an error saying the number was too large. This is beginning to get interesting...is it really possible to represent bits as an integer while preserving the bits? I need to convert the bits 01010010100101 to an int that equals 01010010100101.

Administrator · Answer

64 bit integer will be good I think. If no try to use string: Int64 longInt=Int64.Parse(...)

Administrator · Answer

Just tried it and the resulting integer is something like 252569461. I need to preserve the bits but represent them as an integer; the string 0100101001001 I want converted to the int 0100101001001. Possible?

Administrator · Answer

Use something like Convert.ToInt32(myTextBox.Text2)