Evangelista Sami said:hello
how can i know the size of an int in bits?
i looked in limits.h but i did not find anything to get this
i looked in limits.h but i did not find anything to get this
printf("sizeof(int)=%d bits",sizeof(int)*8);
At best, you're going to get an /approximation/ of the number of bits in an int
with...
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
int main(void)
{
printf("An int contains (at most) %d bits\n",sizeof(int) * CHAR_BIT);
return EXIT_SUCCESS;
}
With inspection and proper evaluation of UINT_MAX and INT_MAX, you can deduce
the minimum number of bits allocated to an int.
Should have been ((2^(sizeof(int)*8))+1)Oh wow! Do U meen that ((sizeof(int)*8)+1) not equals to UINT_MAX on any
Witch ofcourse should have been ((2^(sizeof(int)*8))-1) ;-)Should have been ((2^(sizeof(int)*8))+1)
jota said:Witch ofcourse should have been ((2^(sizeof(int)*8))-1) ;-)
Should have been ((2^(sizeof(int)*8))+1)
Lew Pitcher said:Yes, I mean that
(1 << (sizeof(int) * 8)) - 1
is not necessarily equal to UINT_MAX
/possible/)65537. (I didn't say that it is /common/, I just said that it is
/possible/)
Yes! I know my PIC18 compiler UINT_MAX const is
#define UINT_MAX 65535 /* unsigned int */
..but i certenly expect sizeof(int) to be 2 in that case
..ofcourse CHAR_BIT is 8 under that env. but I can certenly se senarios
where it might not be!
Ben said:When CHAR_BIT == 8, that expression is undefined, because
sizeof(int) * 8 will be equal to (or greater than) the width of
an int.
Lew Pitcher said:It is entirely possible to have (sizeof(int)*CHAR_BIT) == 64, but UINT_MAX only
65537. (I didn't say that it is /common/, I just said that it is /possible/)
Ben said:I don't think a UINT_MAX of 65537 is possible, is that just a
typo for 65535?
-----BEGIN PGP SIGNED MESSAGE-----
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Bzzzt. Wrong answer. Thank you for playing.
At best, you're going to get an /approximation/ of the number of bits in an int
with...
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
int main(void)
{
printf("An int contains (at most) %d bits\n",sizeof(int) * CHAR_BIT);
return EXIT_SUCCESS;
-----BEGIN PGP SIGNED MESSAGE-----
Hash: SHA1
Ben, could you explain how the expression is undefined if
CHAR_BIT == 8
and
sizeof(int) == 3
and
UINT_MAX == 66000
?
I think I see your point, but I'd like to be sure.
Jack said:Bzzzt. Wrong answer. Thank you for playing.
Your code above passes a value of type size_t to printf() with a "%d"
conversion specifier. This produces undefined behavior.
Lew Pitcher said:Ben, could you explain how the expression is undefined if
CHAR_BIT == 8
and
sizeof(int) == 3
and
UINT_MAX == 66000
?
Lew Pitcher said:No, it was intended.
I didn't have a real implementation in mind, just something unusual that still
conforms to the standards. AFAIK, the standards don't specify that UINT_MAX must
be some power-of-two minus 1; they simply say that the minimum value for
UINT_MAX is 65535. So, I chose a value for UINT_MAX that is greater than 65535,
but less than 131072.
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