Hi, i right now have a question for my CS Course. is about convert decimal to bi

Question

Hi, i right now have a question for my CS Course. is about convert decimal to binary. we need to modify the code that provide by our professor.


here is the code that provide by my professor:

#include <iostream>
#include <stdlib.h>

using namespace std;

int main(int argc, char *argv[])
{
int x;

// Start with the value 4, whose least significant byte
// is 00000100 in binary
x = 4;
cout << "x = " << x << endl;

// Now we'll "bitwise or" in the value for 2, which is
// 00000010 in binary. The result is 00000110, or 6 (decimal)
x = x | 2;
cout << "x = " << x << endl;

// The notation "x |= y" is equivalent to "x = x | y"
// in C++ (and C). We'll bitwise or a 16 in there, to produce
// 00010110 binary, or 22 decimal. Notice that if you bitwise
// or powers of 2 (4, 2, 16 so far), this is the same as adding
// them.
x |= 16;
cout << "x = " << x << endl;

// Notice that when you bitwise or a 16 again, the result is
// still 22 (this is an "or", not an xor)
x |= 16;
cout << "x = " << x << endl;

// Let's work with the "bitwise and" operator
// x &= 21 will show which bits x and 20 BOTH have set in their
// binary representation.
// x's current value is 22, which in binary is 00010110
// 21 in binary is 00010101
// the bits they have in common are 00010100
// the decimal value of 00010100 is 4 + 16 = 20
x &= 21;
cout << "x = " << x << endl;

// Notice (for your Lab) that if you bitwise & with a value
// that is a power of two, you can check if that particular
// bit is set.
// Let's see if 427 has the "twos" bit set
if (427 & 2)
cout << "427 does have the 2's bit set" << endl;
else
cout << "427 does NOT have the 2's bit set" << endl;

// Notice above that we're using the fact that any non-zero value
// for the expression (427 & 2) will make it true. Equivalently,
// we could have written "if ((427 & 2) == 2)"
// Let's do that to see if the ones bit is set:
if ((427 & 4) == 4)
cout << "427 does have the 4's bit set" << endl;
else
cout << "427 does NOT have the 4's bit set" << endl;

// Finally, let's see how the shift operator works. Shifting
// moves bits one way or the other. Let's start with 1 and shift
// in from the right, 31 times, to list the powers of two.
int i;
int counter;
counter = 1;
for (i=1; i<32; i++) {
cout << "2^" << i << " = " << counter << endl;
counter <<= 1;
}
}

Explanation / Answer

please rate - thanks

you didn't say what to modify it to

this converts decimal to binary using shifting and anding

#include <stdlib.h>
#include <iostream>
using namespace std;
int main(int argc, char *argv[])
{
int num,shifted,i;
cout<<"Enter integer you want to convert (0 when done): ";
cin>>num;
while(num!=0)
   {cout<<"bin: ";
     for(i=31;i>=0;i--)
       {shifted=num;           //don't destroy original number
        shifted=shifted>>i;    //shift to the right
        shifted=shifted&1;    //use mask to get right most bit
        cout<<shifted;     //print the digit
        }
       cout<<endl<<endl;        //when done with all digits skip 2 lines
    cout<<"Enter integer you want to convert (0 when done): ";
    cin>>num;            
   }
system("pause");
return 0;  
}

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