Convert to binary base long double

[jose luis fernandez diaz]

Hi,

The program below prints a unsigned short in binary base:
#include <iostream>

int main() {
unsigned short mask =0x1;
unsigned short us =0x0071;

mask<<=sizeof(u nsigned short)*8-1;
for (int i = 0; i < sizeof(unsigned short)*8; i++)
{
cout << ((us & mask)? 1:0);
us <<= 1;
}

return 0;
}
How can I print the binary base of a long double ?
Thanks,
Jose Luis.

[Asfand Yar Qazi]

jose luis fernandez diaz wrote:

Hi,

The program below prints a unsigned short in binary base:
#include <iostream>

int main() {
unsigned short mask =0x1;
unsigned short us =0x0071;

mask<<=sizeof(u nsigned short)*8-1;
for (int i = 0; i < sizeof(unsigned short)*8; i++)
{
cout << ((us & mask)? 1:0);
us <<= 1;
}

return 0;
}
How can I print the binary base of a long double ?

I think long double should be 64 bits long... so cast to a 'long long*'
(if your compiler supports it) and away you go as above. I think :-)

--
http://www.it-is-truth.org/

[Gianni Mariani]

jose luis fernandez diaz wrote:

Hi,

The program below prints a unsigned short in binary base:
#include <iostream>

int main() {
unsigned short mask =0x1;
unsigned short us =0x0071;

mask<<=sizeof(u nsigned short)*8-1;
for (int i = 0; i < sizeof(unsigned short)*8; i++)
{
cout << ((us & mask)? 1:0);
us <<= 1;
}

return 0;
}
How can I print the binary base of a long double ?

You need to know the format of floating point types for your platform.
Most modern systems use the IEEE floating point formats.

[David Harmon]

On 23 Jan 2004 01:48:22 -0800 in comp.lang.c++,
jo************* *********@yahoo .es (jose luis fernandez diaz) was alleged
to have written:

How can I print the binary base of a long double ?

The very thought of doing such a low-level and implementation
specific thing is naughty. Nevertheless...

Create a union type containing a long double and
array of char[sizeof(long double}]. Assign to the long
double member, then print the chars one by one. This yields
explicitly "undefined behavior", but has been known to work
for some implementation.

[Kevin Goodsell]

Asfand Yar Qazi wrote:

I think long double should be 64 bits long... so cast to a 'long long*'
(if your compiler supports it) and away you go as above. I think :-)

That is a very non-portable suggestion. Just performing the cast may
give undefined behavior.

-Kevin
--
My email address is valid, but changes periodically.
To contact me please use the address from a recent posting.

[Kevin Goodsell]

David Harmon wrote:

On 23 Jan 2004 01:48:22 -0800 in comp.lang.c++,
jo************* *********@yahoo .es (jose luis fernandez diaz) was alleged
to have written:

How can I print the binary base of a long double ?

The very thought of doing such a low-level and implementation
specific thing is naughty. Nevertheless...

Create a union type containing a long double and
array of char[sizeof(long double}]. Assign to the long
double member, then print the chars one by one. This yields
explicitly "undefined behavior", but has been known to work
for some implementation.

Why use undefined behavior and risk complete failure when there's a
simple way to do it in a well-defined manner?

long double val = 12.34;
unsigned char bytes[sizeof(long double)];
memcpy(bytes, &val, sizeof(val));

Now 'bytes' contains the object-representation (C99 term for it) of val,
and you can print it out however you see fit.

-Kevin
--
My email address is valid, but changes periodically.
To contact me please use the address from a recent posting.

[Jack Klein]

On 23 Jan 2004 01:48:22 -0800, jo************* *********@yahoo .es (jose
luis fernandez diaz) wrote in comp.lang.c++:

Hi,

The program below prints a unsigned short in binary base:
#include <iostream>

int main() {
unsigned short mask =0x1;
unsigned short us =0x0071;

mask<<=sizeof(u nsigned short)*8-1;
This is hideous and makes non-portable assumptions. Include <climits>
and define mask like this:

unsigned int mask = USHRT_MAX - (USHRT_MAX >> 1);
for (int i = 0; i < sizeof(unsigned short)*8; i++)
Replace this with:

for ( ; max != 0; max >> 1)
{
cout << ((us & mask)? 1:0);
us <<= 1;
Eliminate the line above.
}

return 0;
}

--
Jack Klein
Home: http://JK-Technology.Com
FAQs for
comp.lang.c http://www.eskimo.com/~scs/C-faq/top.html
comp.lang.c++ http://www.parashift.com/c++-faq-lite/
alt.comp.lang.l earn.c-c++
http://www.contrib.andrew.cmu.edu/~a...FAQ-acllc.html

[Old Wolf]

Jack Klein <ja*******@spam cop.net> wrote in message news:<he******* *************** **********@4ax. com>...

This is hideous and makes non-portable assumptions. Include <climits>
and define mask like this:

unsigned int mask = USHRT_MAX - (USHRT_MAX >> 1);

for (int i = 0; i < sizeof(unsigned short)*8; i++)
Replace this with:

for ( ; max != 0; max >> 1)

for ( ; mask != 0; mask >>= 1)

{
cout << ((us & mask)? 1:0);
us <<= 1;

Eliminate the line above.

}

This has caught me many a time :) (especially on compilers
that fail to give it a diagnostic)