Can anybody give an example of c++ code that can easily convert a decimal value to binary and a binary value to decimal please?
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In what format? Strings? Numerical types don't care about base, since they're all stored as binary data regardless... – Amber Mar 30 '10 at 20:04
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8Where'd the `"guru"` part in the name come from? ;-) – mjv Mar 30 '10 at 20:04
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it can be anything int...String...i was looking for smallest and efficient code. – Jony Mar 30 '10 at 20:06
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1Likely duplicate: http://stackoverflow.com/questions/819487/efficiently-convert-between-hex-binary-and-decimal-in-c-c – mjv Mar 30 '10 at 20:06
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The question doesn't make sense if it's an `int`, Codeguru. – Amber Mar 30 '10 at 20:07
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@Dav: I'm guessing Codeguru actually means that the input could be anything, but the output should be a string? – Cascabel Mar 30 '10 at 20:08
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1@mjv: +1. My handle used to be UfcChampion9YearsRunning... – Mar 30 '10 at 20:08
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@MaxGuernseyIII: LOL. Assuming UFC is as in "Ultimate Fighting", at least nobody would dare make fun of your handle... I couldn't resist here with our OP, a guru who doesn't find his way around google or a good reference book ;-) – mjv Mar 30 '10 at 20:28
4 Answers
Well, your question is really vague, so this answer is the same.
string DecToBin(int number)
{
if ( number == 0 ) return "0";
if ( number == 1 ) return "1";
if ( number % 2 == 0 )
return DecToBin(number / 2) + "0";
else
return DecToBin(number / 2) + "1";
}
int BinToDec(string number)
{
int result = 0, pow = 1;
for ( int i = number.length() - 1; i >= 0; --i, pow <<= 1 )
result += (number[i] - '0') * pow;
return result;
}
You should check for overflow and do input validation of course.
x << 1 == x * 2
Here's a way to convert to binary that uses a more "programming-like" approach rather than a "math-like" approach, for lack of a better description (the two are actually identical though, since this one just replaces divisions by right shifts, modulo by a bitwise and, recursion with a loop. It's kind of another way of thinking about it though, since this makes it obvious you are extracting the individual bits).
string DecToBin2(int number)
{
string result = "";
do
{
if ( (number & 1) == 0 )
result += "0";
else
result += "1";
number >>= 1;
} while ( number );
reverse(result.begin(), result.end());
return result;
}
And here is how to do the conversion on paper:

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What if we want to have a fixed length of binary number? For example, if we would like to have an 8 bit representation, then `1` has to be returned as `00000001`. – Rasoul Sep 08 '13 at 16:10
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@Rasoul - then just add leading zeros until you get your desired length. Some programming languages provide methods to do this out of the box. – IVlad Sep 08 '13 at 16:13
//The shortest solution to convert dec to bin in c++
void dec2bin(int a) {
if(a!=0) dec2bin(a/2);
if(a!=0) cout<<a%2;
}
int main() {
int a;
cout<<"Enter the number: "<<endl;
cin>>a;
dec2bin(a);
return 0;
}

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strtol
will convert a binary string like "011101" to an internal value (which will normally be stored in binary as well, but you don't need to worry much about that). A normal conversion (e.g. operator<<
with std:cout
) will give the same value in decimal.

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I assume you want a string to binary conversion?
template<typename T> T stringTo( const std::string& s )
{
std::istringstream iss(s);
T x;
iss >> x;
return x;
};
template<typename T> inline std::string toString( const T& x )
{
std::ostringstream o;
o << x;
return o.str();
}
use these like this:
int x = 32;
std:string decimal = toString<int>(x);
int y = stringTo<int>(decimal);

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