Hi!
How can I output value of char or int in binary form with printf(); ?
thanx in advance
26  50307 
Carramba wrote:
Hi!
How can I output value of char or int in binary form with printf(); ?
thanx in advance
There is no standard format specifier for binary form. You will have
to do the conversion manually, testing each bit from highest to
lowest, printing '0' if it's not set, and '1' if it is.
On Sat, 24 Mar 2007 08:55:36 +0100, Carramba <us**@example.netwrote:
>Hi!
How can I output value of char or int in binary form with printf(); ?
thanx in advance
The C Standards do not define a conversion specifier for printf() to
output in binary. The only portable way to do this is to roll your
own. Here's a start:
printf("%s\n", int_to_binary_string(my_int));
Make sure when you implement int_to_binary_string() that it works with
most desktop targets where sizeof(int) * CHAR_BIT = 32 as well on many
embedded targets where sizeof(int) * CHAR_BIT = 16.
Best regards
--
jay
"Carramba" <us**@example.netwrote in message
news:46*********************@news.luth.se...
Hi!
How can I output value of char or int in binary form with printf(); ?
thanx in advance
#include <limits.h>
/*
convert machine number to human-readable binary string.
Returns: pointer to static string overwritten with each call.
*/
char *itob(int x)
{
static char buff[sizeof(int) * CHAR_BIT + 1];
int i;
int j = sizeof(int) * CHAR_BIT - 1;
buff[j] = 0;
for(i=0;i<sizeof(int) * CHAR_BIT; i++)
{
if(x & (1 << i))
buff[j] = '1';
else
buff[j] = '0';
j--;
}
return buff;
}
Call
int x = 100;
printf("%s", itob(x));
You might want something more elaborate to cut leading zeroes or handle
negative numbers.
Harald van Dijk wrote:
Carramba wrote:
>Hi!
How can I output value of char or int in binary form with printf(); ?
thanx in advance
There is no standard format specifier for binary form. You will have
to do the conversion manually, testing each bit from highest to
lowest, printing '0' if it's not set, and '1' if it is.
thanx, maybe you have so suggestion or link for further reading on how
to do it?
thanx ! have few questions about this code :)
Malcolm McLean wrote:
>
"Carramba" <us**@example.netwrote in message
news:46*********************@news.luth.se...
>Hi!
How can I output value of char or int in binary form with printf(); ?
thanx in advance
#include <limits.h>
/*
convert machine number to human-readable binary string.
Returns: pointer to static string overwritten with each call.
*/
char *itob(int x)
{
static char buff[sizeof(int) * CHAR_BIT + 1];
why sizeof(int) * CHAR_BIT + 1 ? what does it mean?
int i;
int j = sizeof(int) * CHAR_BIT - 1;
why sizeof(int) * CHAR_BIT - 1 ? what does it mean?
>
buff[j] = 0;
for(i=0;i<sizeof(int) * CHAR_BIT; i++)
{
if(x & (1 << i))
buff[j] = '1';
else
buff[j] = '0';
j--;
}
return buff;
}
Call
int x = 100;
printf("%s", itob(x));
You might want something more elaborate to cut leading zeroes or handle
negative numbers.
Carramba wrote:
Harald van Dijk wrote:
Carramba wrote:
Hi!
How can I output value of char or int in binary form with printf(); ?
thanx in advance
There is no standard format specifier for binary form. You will have
to do the conversion manually, testing each bit from highest to
lowest, printing '0' if it's not set, and '1' if it is.
thanx, maybe you have so suggestion or link for further reading on how
to do it?
Others have given code already, but here's mine anyway:
#include <limits.h>
#include <stdio.h>
void print_char_binary(char val)
{
char mask;
if(CHAR_MIN < 0)
{
if(val < 0
|| val == 0 && val & CHAR_MAX)
putchar('1');
else
putchar('0');
}
for(mask = (CHAR_MAX >1) + 1; mask != 0; mask >>= 1)
if(val & mask)
putchar('1');
else
putchar('0');
}
void print_int_binary(int val)
{
int mask;
if(val < 0
|| val == 0 && val & INT_MAX)
putchar('1');
else
putchar('0');
for(mask = (INT_MAX >1) + 1; mask != 0; mask >>= 1)
if(val & mask)
putchar('1');
else
putchar('0');
}
Carramba wrote:
>
Hi!
How can I output value of char or int in binary form with printf(); ?
thanx in advance
/* BEGIN output from new.c */
1 = 00000001
2 = 00000010
3 = 00000011
4 = 00000100
5 = 00000101
6 = 00000110
7 = 00000111
8 = 00001000
9 = 00001001
10 = 00001010
11 = 00001011
12 = 00001100
13 = 00001101
14 = 00001110
15 = 00001111
16 = 00010000
17 = 00010001
18 = 00010010
19 = 00010011
20 = 00010100
/* END output from new.c */
/* BEGIN new.c */
#include <limits.h>
#include <stdio.h>
#define STRING "%2d = %s\n"
#define E_TYPE char
#define P_TYPE int
#define INITIAL 1
#define FINAL 20
#define INC(E) (++(E))
typedef E_TYPE e_type;
typedef P_TYPE p_type;
void bitstr(char *str, const void *obj, size_t n);
int main(void)
{
e_type e;
char ebits[CHAR_BIT * sizeof e + 1];
puts("\n/* BEGIN output from new.c */\n");
e = INITIAL;
bitstr(ebits, &e, sizeof e);
printf(STRING, (p_type)e, ebits);
while (FINAL e) {
INC(e);
bitstr(ebits, &e, sizeof e);
printf(STRING, (p_type)e, ebits);
}
puts("\n/* END output from new.c */");
return 0;
}
void bitstr(char *str, const void *obj, size_t n)
{
unsigned mask;
const unsigned char *byte = obj;
while (n-- != 0) {
mask = ((unsigned char)-1 >1) + 1;
do {
*str++ = (char)(mask & byte[n] ? '1' : '0');
mask >>= 1;
} while (mask != 0);
}
*str = '\0';
}
/* END new.c */
--
pete
"Carramba" <us**@example.netschrieb im Newsbeitrag
news:46*********************@news.luth.se...
thanx ! have few questions about this code :)
Malcolm McLean wrote:
>>
"Carramba" <us**@example.netwrote in message
news:46*********************@news.luth.se...
>>Hi!
How can I output value of char or int in binary form with printf(); ?
thanx in advance
#include <limits.h>
/*
convert machine number to human-readable binary string.
Returns: pointer to static string overwritten with each call.
*/
char *itob(int x)
{
static char buff[sizeof(int) * CHAR_BIT + 1];
why sizeof(int) * CHAR_BIT + 1 ? what does it mean?
If you want to put an in't binary representation ionto a string you need
that much space.
On many implementations sizeof(int) is 4 and CHAR_BIT is 8, so you'd need an
array of 33 chars (including the teminating null byte).
I'd use sizeof(x) instead of sizeof(int), that way you can easily change the
function to work on e.g. long long
> int i;
int j = sizeof(int) * CHAR_BIT - 1;
why sizeof(int) * CHAR_BIT - 1 ? what does it mean?
Arrays count from 0 to n and the terminating null byte isn't needed , so the
index goes from 0 to 31 (assuming the same sizes as above)
> buff[j] = 0;
for(i=0;i<sizeof(int) * CHAR_BIT; i++)
{
if(x & (1 << i))
buff[j] = '1';
else
buff[j] = '0';
j--;
}
return buff;
}
Call
int x = 100;
printf("%s", itob(x));
You might want something more elaborate to cut leading zeroes or handle
negative numbers
Bye, Jojo.
Malcolm McLean wrote:
for(i=0;i<sizeof(int) * CHAR_BIT; i++)
{
if(x & (1 << i))
There are some problems with that shift expression.
(1 << sizeof(int) * CHAR_BIT - 1) is undefined.
--
pete
"pete" <pf*****@mindspring.comwrote in message
Malcolm McLean wrote:
> for(i=0;i<sizeof(int) * CHAR_BIT; i++)
{
if(x & (1 << i))
There are some problems with that shift expression.
(1 << sizeof(int) * CHAR_BIT - 1) is undefined.
The function should take an unsigned int. However I didn't want to add that
complication for the OP. It should work OK on almost every platform.
--
Free games and programming goodies. http://www.personal.leeds.ac.uk/~bgy1mm
Malcolm McLean wrote:
>
"pete" <pf*****@mindspring.comwrote in message
(1 << sizeof(int) * CHAR_BIT - 1) is undefined.
The function should take an unsigned int.
That makes no difference.
The evaluation of (1 << sizeof(int) * CHAR_BIT - 1)
in a program is always undefined,
and prevents a program from being a "correct program".
(1 << sizeof(int) * CHAR_BIT - 1) can't be a positive value.
However I didn't want to add that
complication for the OP.
That expression would be perfect to use as
an example of how not to write code.
It should work OK on almost every platform.
(1u << sizeof(int) * CHAR_BIT - 1) is defined.
Your initial value of j is also wrong:
int j = sizeof(int) * CHAR_BIT - 1;
buff[j] = 0;
for(i=0;i<sizeof(int) * CHAR_BIT; i++)
{
if(x & (1 << i))
buff[j] = '1';
else
buff[j] = '0';
As you can see in your code above,
the first side effect of the for loop,
is to overwrite the null terminator.
/* BEGIN new.c */
#include <stdio.h>
#include <limits.h>
char *itob(unsigned x);
int main(void)
{
printf("%s\n", itob(100));
return 0;
}
char *itob(unsigned x)
{
unsigned i;
unsigned j;
static char buff[sizeof x * CHAR_BIT + 1];
j = sizeof x * CHAR_BIT;
buff[j--] = '\0';
for (i = 0; i < sizeof x * CHAR_BIT; i++) {
if (x & (1u << i)) {
buff[j--] = '1';
} else {
buff[j--] = '0';
}
if ((1u << i) == UINT_MAX / 2 + 1) {
break;
}
}
while (i++ < sizeof x * CHAR_BIT) {
buff[j--] = '0';
}
return buff;
}
/* END new.c */
--
pete
"pete" <pf******@mindspring.comwrote in message
Malcolm McLean wrote:
>>
"pete" <pf*****@mindspring.comwrote in message
(1 << sizeof(int) * CHAR_BIT - 1) is undefined.
The function should take an unsigned int.
That makes no difference.
The evaluation of (1 << sizeof(int) * CHAR_BIT - 1)
in a program is always undefined,
and prevents a program from being a "correct program".
(1 << sizeof(int) * CHAR_BIT - 1) can't be a positive value.
>However I didn't want to add that
complication for the OP.
That expression would be perfect to use as
an example of how not to write code.
>It should work OK on almost every platform.
(1u << sizeof(int) * CHAR_BIT - 1) is defined.
Your initial value of j is also wrong:
int j = sizeof(int) * CHAR_BIT - 1;
buff[j] = 0;
for(i=0;i<sizeof(int) * CHAR_BIT; i++)
{
if(x & (1 << i))
buff[j] = '1';
else
buff[j] = '0';
As you can see in your code above,
the first side effect of the for loop,
is to overwrite the null terminator.
/* BEGIN new.c */
#include <stdio.h>
#include <limits.h>
char *itob(unsigned x);
int main(void)
{
printf("%s\n", itob(100));
return 0;
}
char *itob(unsigned x)
{
unsigned i;
unsigned j;
static char buff[sizeof x * CHAR_BIT + 1];
j = sizeof x * CHAR_BIT;
buff[j--] = '\0';
for (i = 0; i < sizeof x * CHAR_BIT; i++) {
if (x & (1u << i)) {
buff[j--] = '1';
} else {
buff[j--] = '0';
}
if ((1u << i) == UINT_MAX / 2 + 1) {
break;
}
}
while (i++ < sizeof x * CHAR_BIT) {
buff[j--] = '0';
}
return buff;
}
/* END new.c */
unsigned integers aren't allowed padding bits so you don't need all that
complication.
A pathological platform might break on the expression 1 << int bits - 1,
agreed. To be strictly correct we need to do the calculations in unsigned
integers, but I've explained why I didn't do that.
The off by one error in writing the nul was a slip. Of course I didn't
realise because the static array was zero intilaised anyway. So well
spotted.
--
Free games and programming goodies. http://www.personal.leeds.ac.uk/~bgy1mm
Malcolm McLean wrote:
unsigned integers aren't allowed padding bits
Wrong again.
unsigned char isn't allowed padding bits.
UINT_MAX is allowed to be as low as INT_MAX,
and you can't achieve that without padding bits
in the unsigned int type.
--
pete
Carramba wrote:
Harald van Dijk wrote:
>Carramba wrote:
>>Hi!
How can I output value of char or int in binary form with printf(); ?
thanx in advance
There is no standard format specifier for binary form. You will have
to do the conversion manually, testing each bit from highest to
lowest, printing '0' if it's not set, and '1' if it is.
thanx, maybe you have so suggestion or link for further reading on how
to do it?
Think about it!
void bits(unsigned char b, int n) {
for (--n; n >= 0; --n)
putchar((b & 1 << n) ? '1' : '0');
putchar(' ');
}
Now if you call it..
bits(195, 8);
...you'll get '11000011 ' on the stdout stream.
--
Joe Wright
"Everything should be made as simple as possible, but not simpler."
--- Albert Einstein ---
On 24 Mar 2007 04:42:24 -0700, "Harald van D?k" <tr*****@gmail.com>
wrote:
>Carramba wrote:
>Harald van D?k wrote:
Carramba wrote:
Hi!
How can I output value of char or int in binary form with printf(); ?
thanx in advance
There is no standard format specifier for binary form. You will have
to do the conversion manually, testing each bit from highest to
lowest, printing '0' if it's not set, and '1' if it is.
thanx, maybe you have so suggestion or link for further reading on how
to do it?
Others have given code already, but here's mine anyway:
#include <limits.h>
#include <stdio.h>
void print_char_binary(char val)
{
char mask;
if(CHAR_MIN < 0)
{
if(val < 0
|| val == 0 && val & CHAR_MAX)
When will the expression following the && evaluate to 1? Is it
something to do with ones complement or signed magnitude
representations?
putchar('1');
else
putchar('0');
}
for(mask = (CHAR_MAX >1) + 1; mask != 0; mask >>= 1)
Is it a requirement that (CHAR_MAX>>1)+1 be a power of 2? It is a
requirement for UCHAR_MAX but what if char is signed? (If CHAR_BIT is
9, could SCHAR_MAX and CHAR_MAX be 173?)
if(val & mask)
putchar('1');
else
putchar('0');
}
void print_int_binary(int val)
{
int mask;
if(val < 0
|| val == 0 && val & INT_MAX)
putchar('1');
else
putchar('0');
for(mask = (INT_MAX >1) + 1; mask != 0; mask >>= 1)
There does not appear to be a similar requirement for INT_MAX either.
if(val & mask)
putchar('1');
else
putchar('0');
}
Remove del for email
Barry Schwarz wrote:
On 24 Mar 2007 04:42:24 -0700, "Harald van D?k" <tr*****@gmail.com>
wrote:
Carramba wrote:
Harald van D?k wrote:
Carramba wrote:
Hi!
How can I output value of char or int in binary form with printf(); ?
thanx in advance
There is no standard format specifier for binary form. You will have
to do the conversion manually, testing each bit from highest to
lowest, printing '0' if it's not set, and '1' if it is.
thanx, maybe you have so suggestion or link for further reading on how
to do it?
Others have given code already, but here's mine anyway:
#include <limits.h>
#include <stdio.h>
void print_char_binary(char val)
{
char mask;
if(CHAR_MIN < 0)
{
if(val < 0
|| val == 0 && val & CHAR_MAX)
When will the expression following the && evaluate to 1? Is it
something to do with ones complement or signed magnitude
representations?
It accounts for ones' complement, where all bits 1 is a possible
representation of 0.
It does not account for sign and magnitude, where all value bits 0 and
sign bit 1 is a representation of 0. This will be printed as all bits
zero, which is a different representation of the same value.
putchar('1');
else
putchar('0');
}
for(mask = (CHAR_MAX >1) + 1; mask != 0; mask >>= 1)
Is it a requirement that (CHAR_MAX>>1)+1 be a power of 2? It is a
requirement for UCHAR_MAX but what if char is signed? (If CHAR_BIT is
9, could SCHAR_MAX and CHAR_MAX be 173?)
[ And a similar comment for INT_MAX snipped ]
The only allowed representation systems for signed integer types are
two's complement, ones' complement, and sign and magnitude. All three
have the maximum value as a power of two minus one. (IIRC, this is new
in C99, but it was added because there were no other systems even
though C90 allowed it.)
"Harald van Dijk" <tr*****@gmail.comwrote in message
>
The only allowed representation systems for signed integer types are
two's complement, ones' complement, and sign and magnitude. All three
have the maximum value as a power of two minus one. (IIRC, this is new
in C99, but it was added because there were no other systems even
though C90 allowed it.)
That's typical committee thinking. No engineer is going to devise a new
method of representing integers for the fun of it, but because there is some
technical advantage or requirement. At which point the standard becomes a
dead letter. If the super-whizzy-fibby machine needs Fibonacci
representation for its quantum coherence modulator unit, the either C can't
be used on such a machine or the rule will change. So it is a completely
pointless regulation.
--
Free games and programming goodies. http://www.personal.leeds.ac.uk/~bgy1mm
On 24 Mar 2007 10:51:58 -0700, "Harald van D?k" <tr*****@gmail.com>
wrote:
>Barry Schwarz wrote:
>On 24 Mar 2007 04:42:24 -0700, "Harald van D?k" <tr*****@gmail.com>
wrote:
>Carramba wrote:
Harald van D?k wrote:
Carramba wrote:
Hi!
How can I output value of char or int in binary form with printf(); ?
thanx in advance
There is no standard format specifier for binary form. You will have
to do the conversion manually, testing each bit from highest to
lowest, printing '0' if it's not set, and '1' if it is.
thanx, maybe you have so suggestion or link for further reading on how
to do it?
Others have given code already, but here's mine anyway:
#include <limits.h>
#include <stdio.h>
void print_char_binary(char val)
{
char mask;
if(CHAR_MIN < 0)
{
if(val < 0
|| val == 0 && val & CHAR_MAX)
When will the expression following the && evaluate to 1? Is it
something to do with ones complement or signed magnitude
representations?
It accounts for ones' complement, where all bits 1 is a possible
representation of 0.
It does not account for sign and magnitude, where all value bits 0 and
sign bit 1 is a representation of 0. This will be printed as all bits
zero, which is a different representation of the same value.
putchar('1');
else
putchar('0');
}
for(mask = (CHAR_MAX >1) + 1; mask != 0; mask >>= 1)
Is it a requirement that (CHAR_MAX>>1)+1 be a power of 2? It is a
requirement for UCHAR_MAX but what if char is signed? (If CHAR_BIT is
9, could SCHAR_MAX and CHAR_MAX be 173?)
[ And a similar comment for INT_MAX snipped ]
The only allowed representation systems for signed integer types are
two's complement, ones' complement, and sign and magnitude. All three
have the maximum value as a power of two minus one. (IIRC, this is new
in C99, but it was added because there were no other systems even
though C90 allowed it.)
n1124 says that UCHAR_MAX must be equal to 2^CHAR_BIT-1 which I
mentioned in my question. For SCHAR_MAX, there is no such
requirement. It is required to be at least (minimum value) 127 which
is 2^7-1 but for larger values of CHAR_BIT there is no additional
restriction. Again, if CHAR_BIT is 9, could SCHAR_MAX and CHAR_MAX be
173?
Remove del for email
Barry Schwarz wrote:
On 24 Mar 2007 10:51:58 -0700, "Harald van D?k" <tr*****@gmail.com>
wrote:
The only allowed representation systems for signed integer types are
two's complement, ones' complement, and sign and magnitude. All three
have the maximum value as a power of two minus one. (IIRC, this is new
in C99, but it was added because there were no other systems even
though C90 allowed it.)
n1124 says that UCHAR_MAX must be equal to 2^CHAR_BIT-1 which I
mentioned in my question. For SCHAR_MAX, there is no such
requirement. It is required to be at least (minimum value) 127 which
is 2^7-1 but for larger values of CHAR_BIT there is no additional
restriction. Again, if CHAR_BIT is 9, could SCHAR_MAX and CHAR_MAX be
173?
Again, no. The only allowed representation systems for signed integers
are those three. If SCHAR_MAX is 173 (or if INT_MAX is 99999), then
the representation system cannot be one of those three, so the
implementation would violate 6.2.6.2p2. The fact that there is an
explicit statement that UCHAR_MAX must equal 2^CHAR_BIT - 1 doesn't
seem relevant to me, because unsigned char is already a special case
(as the only integer type that may not contain trap representations or
padding bits).
Carramba <us**@example.netwrote:
# Hi!
#
# How can I output value of char or int in binary form with printf(); ?
You might sprintf with %x and then just map the hex digits
to four character strings.
--
SM Ryan http://www.rawbw.com/~wyrmwif/
Quit killing people. That's high profile.
On 24 Mar 2007 14:39:23 -0700, "Harald van D?k" <tr*****@gmail.com>
wrote:
>Barry Schwarz wrote:
>On 24 Mar 2007 10:51:58 -0700, "Harald van D?k" <tr*****@gmail.com>
wrote:
>The only allowed representation systems for signed integer types are
two's complement, ones' complement, and sign and magnitude. All three
have the maximum value as a power of two minus one. (IIRC, this is new
in C99, but it was added because there were no other systems even
though C90 allowed it.)
n1124 says that UCHAR_MAX must be equal to 2^CHAR_BIT-1 which I
mentioned in my question. For SCHAR_MAX, there is no such
requirement. It is required to be at least (minimum value) 127 which
is 2^7-1 but for larger values of CHAR_BIT there is no additional
restriction. Again, if CHAR_BIT is 9, could SCHAR_MAX and CHAR_MAX be
173?
Again, no. The only allowed representation systems for signed integers
are those three. If SCHAR_MAX is 173 (or if INT_MAX is 99999), then
the representation system cannot be one of those three, so the
implementation would violate 6.2.6.2p2. The fact that there is an
explicit statement that UCHAR_MAX must equal 2^CHAR_BIT - 1 doesn't
seem relevant to me, because unsigned char is already a special case
(as the only integer type that may not contain trap representations or
padding bits).
I can find no requirement that every possible bit combination must be
a valid value. The fact that a signed 9-bit char can support a value
larger than 173 in any of the three allowed representations doesn't
mean it has to.
Remove del for email
Carramba wrote:
>
How can I output value of char or int in binary form with printf(); ?
#include <stdio.h>
#include <stdlib.h>
static void binprt(long i) {
if (i / 2) binprt(i / 2);
putchar('0' + i % 2);
} /* binprt */
/* ----------------- */
int main(int argc, char* *argv) {
long x;
if ((argc != 2) || (1 != sscanf(argv[1], "%ld", &x))) {
fprintf(stderr, "Usage: binprt value\n");
exit(EXIT_FAILURE);
}
binprt(x);
putchar('\n');
return 0;
}
--
Chuck F (cbfalconer at maineline dot net)
Available for consulting/temporary embedded and systems.
<http://cbfalconer.home.att.net>
--
Posted via a free Usenet account from http://www.teranews.com
Barry Schwarz wrote:
I can find no requirement that every possible bit combination must be
a valid value. The fact that a signed 9-bit char can support a value
larger than 173 in any of the three allowed representations doesn't
mean it has to.
There are not three allowable representations for positive values
of signed types.
N869
6.2.6.2 Integer types
[#2] For signed integer types, the bits of the object
representation shall be divided into three groups: value
bits, padding bits, and the sign bit. There need not be any
padding bits; there shall be exactly one sign bit. Each bit
that is a value bit shall have the same value as the same
bit in the object representation of the corresponding
unsigned type (if there are M value bits in the signed type
and N in the unsigned type, then M<=N). If the sign bit is
zero, it shall not affect the resulting value.
The only description of value bits,
is in the description of the representation of unsigned types,
so, value bits in the signed type
should behave the same way for positive values.
--
pete
Barry Schwarz wrote:
On 24 Mar 2007 14:39:23 -0700, "Harald van D?k" <tr*****@gmail.com>
wrote:
Barry Schwarz wrote:
On 24 Mar 2007 10:51:58 -0700, "Harald van D?k" <tr*****@gmail.com>
wrote:
The only allowed representation systems for signed integer types are
two's complement, ones' complement, and sign and magnitude. All three
have the maximum value as a power of two minus one. (IIRC, this is new
in C99, but it was added because there were no other systems even
though C90 allowed it.)
n1124 says that UCHAR_MAX must be equal to 2^CHAR_BIT-1 which I
mentioned in my question. For SCHAR_MAX, there is no such
requirement. It is required to be at least (minimum value) 127 which
is 2^7-1 but for larger values of CHAR_BIT there is no additional
restriction. Again, if CHAR_BIT is 9, could SCHAR_MAX and CHAR_MAX be
173?
Again, no. The only allowed representation systems for signed integers
are those three. If SCHAR_MAX is 173 (or if INT_MAX is 99999), then
the representation system cannot be one of those three, so the
implementation would violate 6.2.6.2p2. The fact that there is an
explicit statement that UCHAR_MAX must equal 2^CHAR_BIT - 1 doesn't
seem relevant to me, because unsigned char is already a special case
(as the only integer type that may not contain trap representations or
padding bits).
I can find no requirement that every possible bit combination must be
a valid value. The fact that a signed 9-bit char can support a value
larger than 173 in any of the three allowed representations doesn't
mean it has to.
Hmm, okay, there is explicit permission for what would otherwise be
negative zero to be a trap representation, but perhaps that is
redundant too.
The rationale does say that "any result of bitwise manipulation
produces an integer result which can be printed by printf", but since
that is already incorrect for other reasons, it may not apply to 99999
| 16384 either (which would overflow if INT_MAX is 99999).
The following example is from the Users' Reference to B by Ken Thompson:
/* The following function will print a non-negative number, n, to
the base b, where 2<=b<=10, This routine uses the fact that
in the ASCII character set, the digits 0 to 9 have sequential
code values. */
printn(n,b) {
extrn putchar;
auto a;
if(a=n/b) /* assignment, not test for equality */
printn(a, b); /* recursive */
putchar(n%b + '0');
}Simplily adding "void" before "printn", replacing "extrn putchar" with
"#include <stdio.h>" outside the function, and replace "auto" with "int"
translates it into C."Carramba" <us**@example.netha scritto nel messaggio
news:46*********************@news.luth.se...
Hi!
How can I output value of char or int in binary form with printf(); ?
thanx in advance
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