Programming Forums > Application Development > C++

ionexchange · Feb 11th, 2006, 8:52 AM

Hello,
I'm tring to use bitwise operators like <<, |, &, ^ with a double data type.

double foo = 0;
double bar = 0;
double huh = 0;

huh = (foo & bar);

I compile and get an error
invalid operands of type "double" and "double" to binary operator&

I get similar errors when I use | and ^. When I use << I get results as if I'm dealing with two signed long ints. Somewhere in the dusty area of my mind I recall that double is just that, two signed long ints. I need a data type that has 64 bits that I can use these bitwise operators. Is there one out there or do I have to create my own and overload those operators?

Thanks,
Milton

sunnypalsingh · Feb 11th, 2006, 9:58 AM

Can't use doubles/floats with bitwise operators. Use integer types instead.

DaWei · Feb 11th, 2006, 10:40 AM

If you want to mess with a double bitwise, you'll have to cast it to something else. Before you do that, I'd recommend you investigate exactly what a double is and how it's represented in your system, and what endianness is associated with your machine.

ionexchange · Feb 11th, 2006, 10:58 AM

Quote:

Originally Posted by DaWei

If you want to mess with a double bitwise, you'll have to cast it to something else. Before you do that, I'd recommend you investigate exactly what a double is and how it's represented in your system, and what endianness is associated with your machine.

I believe that double is represented as two unsigned long int's, so I guess I'm left with using two unsigned long int's and overloading the operators?

Thanks DaWei and sunnypalsingh for your help.
Milton

DaWei · Feb 11th, 2006, 11:01 AM

Quote:

I believe that double is represented as two unsigned long int's,

You believe wrongly unless you have a reallllllyyyyyy strange machine.

Quote:

Originally Posted by DaWei

I'd recommend you investigate exactly what a double is and how it's represented in your system...

ionexchange · Feb 11th, 2006, 11:35 AM

Okay,
So I'm confused. I determine my machine represents float with big endianness by using the following code.

(Toggle Plain Text)

 #define LITTLE_ENDIAN 0
#define BIG_ENDIAN    1

int main()
{
   float i = 1;
   char *p = (char *) &i;
   if (p[0] == 1) // Lowest address contains the least significant byte
      std::cout << "little" << std::endl;
   else
      std::cout << "big" << std::endl;
return 1;
}

I use

(Toggle Plain Text)

for (unsigned int a = 0; a < 64; a++){
    std::cout << a << " " << ((double)(1<<a)) << std::endl;
}

to determine see how the bits are represented in double. The output for a = 0 through a = 31 is identical for the output for a=32 through a=65. Is there another way to determine what you are refering to investigate exactly what a double is and how it's represented in my system.

Milt

andro · Feb 11th, 2006, 12:32 PM

Take a look at the IEEE Floating point specification, and how float and double's are represented in binary. Then you'll understand.

andro · Feb 11th, 2006, 1:00 PM

Bytes:
31-30 sign
30-23 exponent
23-0 signifcand

(-1)^s * (1 + significand) * 2^(exp-127)

|1|1000 0001|111 0000 0000 0000 0000 0000|

(-1)^1 * (1+.111) * 2^(129-127)
-1 * (1.111) * 2^2
-111.1 (base 2)

=7.5 (base 10)

ionexchange · Feb 11th, 2006, 1:33 PM

Quote:

Originally Posted by andro

Bytes:
31-30 sign
30-23 exponent
23-0 signifcand

(-1)^s * (1 + significand) * 2^(exp-127)

|1|1000 0001|111 0000 0000 0000 0000 0000|

(-1)^1 * (1+.111) * 2^(129-127)
-1 * (1.111) * 2^2
-111.1 (base 2)

=7.5 (base 10)

Thanks, now I understand what I read.

Milt
I have seen the light, and it BURNS!

DaWei · Feb 11th, 2006, 1:42 PM

On my machine, the example given above represents a float. A double is represented in a 64-bit value with a similar structure but a larger number of bits for the mantissa and exponent. This is why you need to investigate for YOUR machine. The IEEE specification is a very good start, and will definitely give you the overall picture, but not all implementations are in accordance with it. You can appreciate that for any given size of binary thangy, only so much information can be encoded. This is why floating point values give you more range, but less precision.

Feb 11th, 2006, 8:52 AM	#1
ionexchange Newbie Join Date: Oct 2005 Posts: 12 Rep Power: 0	bitwise operators and type double Hello, I'm tring to use bitwise operators like <<, \|, &, ^ with a double data type. (Toggle Plain Text) double foo = 0; double bar = 0; double huh = 0; huh = (foo & bar); double foo = 0; double bar = 0; double huh = 0; huh = (foo & bar); I compile and get an error invalid operands of type "double" and "double" to binary operator& I get similar errors when I use \| and ^. When I use << I get results as if I'm dealing with two signed long ints. Somewhere in the dusty area of my mind I recall that double is just that, two signed long ints. I need a data type that has 64 bits that I can use these bitwise operators. Is there one out there or do I have to create my own and overload those operators? Thanks, Milton

Feb 11th, 2006, 9:58 AM	#2
sunnypalsingh Programmer Join Date: Dec 2005 Posts: 33 Rep Power: 0	Can't use doubles/floats with bitwise operators. Use integer types instead. __________________ The only real valuable thing is intuition. -Albert Einstein

Feb 11th, 2006, 10:40 AM	#3
DaWei Resident Grouch Join Date: Jun 2005 Posts: 6,453 Rep Power: 10	If you want to mess with a double bitwise, you'll have to cast it to something else. Before you do that, I'd recommend you investigate exactly what a double is and how it's represented in your system, and what endianness is associated with your machine. __________________ Abstraction doesn't make it impossible to write bad code; it makes it possible to write superior code. Contributor's Corner: Grumpy on C++ Exceptions DaWei on Pointers

Feb 11th, 2006, 11:35 AM	#6
ionexchange Newbie Join Date: Oct 2005 Posts: 12 Rep Power: 0	Okay, So I'm confused. I determine my machine represents float with big endianness by using the following code. (Toggle Plain Text) #define LITTLE_ENDIAN 0 #define BIG_ENDIAN 1 int main() { float i = 1; char p = (char ) &i; if (p[0] == 1) // Lowest address contains the least significant byte std::cout << "little" << std::endl; else std::cout << "big" << std::endl; return 1; } #define LITTLE_ENDIAN 0 #define BIG_ENDIAN 1 int main() { float i = 1; char p = (char ) &i; if (p[0] == 1) // Lowest address contains the least significant byte std::cout << "little" << std::endl; else std::cout << "big" << std::endl; return 1; } I use (Toggle Plain Text) for (unsigned int a = 0; a < 64; a++){ std::cout << a << " " << ((double)(1<<a)) << std::endl; } for (unsigned int a = 0; a < 64; a++){ std::cout << a << " " << ((double)(1<<a)) << std::endl; } to determine see how the bits are represented in double. The output for a = 0 through a = 31 is identical for the output for a=32 through a=65. Is there another way to determine what you are refering to investigate exactly what a double is and how it's represented in my system. Milt

Feb 11th, 2006, 1:42 PM	#10
DaWei Resident Grouch Join Date: Jun 2005 Posts: 6,453 Rep Power: 10	On my machine, the example given above represents a float. A double is represented in a 64-bit value with a similar structure but a larger number of bits for the mantissa and exponent. This is why you need to investigate for YOUR machine. The IEEE specification is a very good start, and will definitely give you the overall picture, but not all implementations are in accordance with it. You can appreciate that for any given size of binary thangy, only so much information can be encoded. This is why floating point values give you more range, but less precision. __________________ Abstraction doesn't make it impossible to write bad code; it makes it possible to write superior code. Contributor's Corner: Grumpy on C++ Exceptions DaWei on Pointers

Feb 11th, 2006, 12:32 PM	#7
andro Professional Programmer Join Date: Oct 2005 Location: California Posts: 294 Rep Power: 3	Take a look at the IEEE Floating point specification, and how float and double's are represented in binary. Then you'll understand.

Feb 11th, 2006, 1:00 PM	#8
andro Professional Programmer Join Date: Oct 2005 Location: California Posts: 294 Rep Power: 3	Bytes: 31-30 sign 30-23 exponent 23-0 signifcand (-1)^s * (1 + significand) * 2^(exp-127) \|1\|1000 0001\|111 0000 0000 0000 0000 0000\| (-1)^1 * (1+.111) * 2^(129-127) -1 * (1.111) * 2^2 -111.1 (base 2) =7.5 (base 10)

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Programming Forums > Application Development > C++

bitwise operators and type double