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(unsigned 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]

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(unsigned 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

 

[Gianni Mariani]

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(unsigned 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]

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]

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

 

[Kevin Goodsell]

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

 

[Jack Klein]

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(unsigned 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.

 

[Old Wolf]

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

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


Replace this with:

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

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

Eliminate the line above.

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