Published by
Jan 6, 2009

A bit of bitwise

Score: 3.4/5 (17 votes)
*****
This is just a little handy structure defining a variable type & a simple operator overload, to get a bit value from a string.

First things first we declare a structure, so:
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struct bin
{

};

Next we need to create a private variable to accept a byte, so we use an unsigned char & let's call it "val":
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private:
unsigned char val;

We need to be able to give the byte a value, so let's create the subroutine to accept the byte & set it to our private variable:
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public:
bin(unsigned char v)
{
val=v;
}

And last, but not least an operator overload so that we can decide which bit we want:
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bool operator[](unsigned char i)
{
return (val&(1<<i))>>i;
}

To quickly explain that above, perhaps complex looking, bitwise statement - it takes a number (i) & then finds if that bit is 1 or 0, then it makes the return value equal to that (1 or 0).

So our completed structure is:
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struct bin
{
private:unsigned char val;
public:bin(unsigned char v){val=v;}
bool operator[](unsigned char i){return (val&(1<<i))>>i;}
};


Now - how to use this, in this example we will take an unsigned char variable called "a" holding the ASCII value of "a", which is 97 - then find all the bits of it.

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int main()
{
unsigned char a=97; // declare our "a"
for(unsigned int c=7;c!=-1;c--) // create a loop to access all the bits
{
printf("%u",bin(a)[c]); // print out the value of a at bit c.
}
getchar(); // wait for user input so we can see previous output
return 0;
}


From this we should get "01100001" - 97 in binary.

I hope this is some help to anyone really, & is my first attempt at an article - any feedback or suggestions are welcome.

Chris.