Bit concatenation in C++

[Manoj7390]

I am implementing turbo encoder, I will give some data as a input and i am converting it into the binary, since i don't know how to store and which data type for binary values, i have stored those binary values as integers only. Now i need to concatenate three bits at a time, I need only 1's and 0's not the integer values. If i shift these values i will get more than 1.

[Manoj7390]

No i dont need the decimal value of binary. I need to that as it is as 1's and 0's. if i have u=1; p=0; q=1; Then i need result = {u,p,q} = "101" (its neither decimal '5' nor decimal 'one hundred one') Just i need to concatenate bits. or else help me how to store binary values and which is the data type for that.

[ThatsAlok]

if you looking for string value as i understand by double quote used by you, you can use sprintf or sprintf_s. now if you want to store the binary value resulting from you operation (as you mentioned they are only 3 bit coming), you can use char datatype (1 byte) and store the decimal equivalent of it there, internally it store its binary equivalent only, for say you got 101 which decimal equivalent is 5 and when you store same in char variable, it binary representation would be 00000100

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[CPallini]

Manoj7390 wrote:

If i shift these values i will get more than 1

Of course. Now look at decimal case: if you 'concatenate' the digits 1,5,7 then you get the number 157 that is no more in the 0-9 range. An alternative would be using strings, that is, you concatenate the character representations of the digits into a string ('1'', '5', '7' => "157"). You may apply exactly the same argument to binary (that is base 2) digits: if you concatenate 1,0,0 you get the number 101, that is decimal 5, no more in the 0-1 range. The string concatenation method would give you "101". You may find interesting the std::bitset[^] class. However it does not prevent you from gaining a proper understanding of numerical bases (see, for instance, Numeral System at Wikipedia[^]) and how numbers are stored in computers.

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[Manoj7390]

But i need to concatenate the bits which are in different variables. how to concatenate those.. Since i cannot use shift operator here because of binary.

[Manoj7390]

Can you please suggest me a program to convert a string into binary, that store in char data type.

[Chris Losinger]

Richard MacCutchan wrote:

If you will not use the shift operators to store the three bits together then you cannot store them.

int Answer = u * 4 + p * 2 + q

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[Chris Losinger]

char * p = "101";
char * pEnd;
int val = strtol(p, &pEnd, 2);
char charval = (char)val;

this may fail if p is more than 7 bits long (since that could overflow a signed char).

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[Lost User]

Which the compiler will almost certainly optimise into shift operations.

Use the best guess

[Lost User]

Manoj7390 wrote:

i cannot use shift operator here because of binary.

What on earth are you talking about? I have shown you exactly how to concatenate these fields into a three bit value. If you really cannot understand that simple operation then I don't think you are going to get very far with this project.

Use the best guess

[Chris Losinger]

possibly. but that extends the definition of "you" quite a bit. int Answer = u + u + u + u + p + p + q;

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[David Crow]

Something like:

int answer = (u*100) + (p*10) + q;

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[Lost User]

Chris Losinger wrote:

that extends the definition of "you" quite a bit.

:confused:

Use the best guess

[Chris Losinger]

you originally wrote "If you will not use the shift operators to store the three bits together then you cannot store them." that's the "you". i think it's reasonable to differentiate between what the programmer writes and what the compiler does.

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[Lost User]

Sorry, but I don't understand what point you are trying to make.

Use the best guess