Exact binary representation of a double

Question

Possible Duplicate:
Float to binary in C++

I have a very small double var, and when I print it I get -0. (using C++). Now in order to get better precision I tried using

cout.precision(18); \\i think 18 is the max precision i can get.
cout.setf(ios::fixed,ios::floatfield);
cout<<var;\\var is a double.

but it just writes -0.00000000000...

I want to see the exact binary representation of the var.

In other words I want to see what binary number is written in the stack memory/register for this var.

Solution 1

union myUnion {
    double dValue;
    uint64_t iValue;
};

myUnion myValue;
myValue.dValue=123.456;
cout << myValue.iValue;

Update:

The version above will work for most purposes, but it assumes 64 bit doubles. This version makes no assumptions and generates a binary representation:

    double someDouble=123.456;
    unsigned char rawBytes[sizeof(double)];

    memcpy(rawBytes,&someDouble,sizeof(double));

    //The C++ standard does not guarantee 8-bit bytes
    unsigned char startMask=1;
    while (0!=static_cast<unsigned char>(startMask<<1)) {
        startMask<<=1;
    }

    bool hasLeadBit=false;   //set this to true if you want to see leading zeros

    size_t byteIndex;
    for (byteIndex=0;byteIndex<sizeof(double);++byteIndex) {
        unsigned char bitMask=startMask;
        while (0!=bitMask) {
            if (0!=(bitMask&rawBytes[byteIndex])) {
                std::cout<<"1";
                hasLeadBit=true;
            } else if (hasLeadBit) {
                std::cout<<"0";
            }
            bitMask>>=1;
        }
    }
    if (!hasLeadBit) {
        std::cout<<"0";
    }

Solution 2

This way is guaranteed to work by the standard:

double d = -0.0;
uint64_t u;
memcpy(&u, &d, sizeof(d));
std::cout << std::hex << u;

Solution 3

Try:

printf("0x%08x\n", myFloat);

This should work for a 32 bit variable, to display it in hex. I've never tried using this technique to see a 64 bit variable, but I think it's:

printf("%016llx\n", myDouble);

EDIT: tested the 64-bit version and it definitely works on Win32 (I seem to recall the need for uppercase LL on GCC.. maybe)

EDIT2: if you really want binary, you are best off using one of the other answers to get a uint64_t version of your double, and then looping:

for ( int i = 63; i >= 0; i-- )
{
    printf( "%d", (myUint64 >> i ) & 1 );
}

Comments

[user690936]

thank you for the response. it does not give a binary number. for the number 2.0 it writes: 0x00000000 it should out put 10, no?

[Hybrid]

I just tried it on my PC and it worked as expected. This code prints in Hexadecimal, not in binary. Also, your comment suggests that you don't understand the difference between an integer representation of 2, and a floating point version. an int with the value 2 has a binary representation of 10, but a float is totally different

[user690936]

i know that there is difference between an integer representation of 2, and a floating point version. but i still need a binary representation. a representation of {0,1}