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How do you invoke the C++ compiler (command line, please)?
You have to g++ (instead of gcc), see, for instance here[^].
Veni, vidi, vici.
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I am using CC - command for compiling, What is the command for g++
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You are using the C compiler.
g++ is the command for invoking the C++ compiler, e.g.
g++ myapp.cpp -o myapp
If you are using a makefile, set CC=g++.
Veni, vidi, vici.
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Manoj7390 wrote: i will send you the code.
No you won't. Just show us the first few lines of your source (where your #include statements are), and the lines that are producing the errors.
Use the best guess
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#include <cstdlib>
#include <iostream>
#include <vector>
using namespace std;
#include <math.h>
Plenty of errors are coming, i couldn't figure it out.
But my question is same program is running in other compilers but why it is not running in g++.
Might be there is a small bug.
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Manoj7390 wrote: Plenty of errors are coming, i couldn't figure it out.
Like I said, show us the line of code and the error message; the first two or three should be enough.
Manoj7390 wrote: Might be there is a small bug.
Most unlikely, it is much more likely that the compiler cannot find one of the include files, or there is some other setting specific to g++ that you need to add. I do not have a Linux system so cannot test this.
Use the best guess
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Thanks.. I have got the output.. I was using CC instead of g++.
It works with g++
Thank you...
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Hi. I have written the program for turbo encoder, for every single bit, it gives three bits as output and I am storing it as a integer. Now the problem is how to concatenate these bits in the integer form.
for example...
int u = 1; int p = 0; int q = 1;
Answer = 101;
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int Answer = (u << 2) | (p << 1) | q
Use the best guess
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I cannot use shifting here, because i am considering bits in the integer form. if i shift, data will change know.
Suggest any other way to store bits in C/C++
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As stated, your sentence makes no sense. Please elaborate.
Veni, vidi, vici.
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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.
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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.
modified 30-May-13 5:30am.
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Manoj7390 wrote: Suggest any other way to store bits in C/C++
This has nothing to do with C/C++, it's how computers work. If you will not use the shift operators to store the three bits together then you cannot store them.
Use the best guess
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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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Which the compiler will almost certainly optimise into shift operations.
Use the best guess
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possibly. but that extends the definition of "you" quite a bit.
int Answer = u + u + u + u + p + p + q;
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Chris Losinger wrote: that extends the definition of "you" quite a bit.
Use the best guess
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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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Sorry, but I don't understand what point you are trying to make.
Use the best guess
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though richard solutions is best.. i am suggesting simpler approach for this in my college days 
int result = i*100 + j *10 + k;
Jokes apart! are you looking for it's integer equivalent. i.e. say you have bits like this i=1,j=0,k=0 for which decimal equivalent is 4 ( binary of 100 = 4 i.e. 22 + 0 + 0 )
"Opinions are neither right nor wrong. I cannot change your opinion. I can, however, change what influences your opinion." - David Crow
Never mind - my own stupidity is the source of every "problem" - Mixture
cheers,
Alok Gupta
VC Forum Q&A :- I/ IV
Support CRY- Child Relief and You
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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.
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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
"Opinions are neither right nor wrong. I cannot change your opinion. I can, however, change what influences your opinion." - David Crow
Never mind - my own stupidity is the source of every "problem" - Mixture
cheers,
Alok Gupta
VC Forum Q&A :- I/ IV
Support CRY- Child Relief and You
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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.
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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
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