Bit concatenation in C++

[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.

Veni, vidi, vici.

[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