What can be assumed about the representation of true?












6















This program returns 0 in my machine:



#include <stdbool.h>

union U {
_Bool b;
char c;
};

int main(void) {
union U u;
u.c = 3;
_Bool b = u.b;
if (b == true) {
return 0;
} else {
return 1;
}
}


AFAICT, _Bool is an integer type that can at least store 0 and 1, and true is the integral constant 1. On my machine, _Bool has a sizeof(_Bool) == 1, and CHAR_BITS == 8, which means that _Bool has 256 representations.



I can't find much in the C standard about the trap representations of _Bool, and I can't find whether creating a _Bool with a representation different from 0 or 1 (on implementations that support more than two representations) is ok, and if it is ok, whether those representations denote true or false.



What I can find in the standard is what happens when a _Bool is compared with an integer, the integer is converted to the 0 representation if it has value 0, and to the 1 representation if it has a value different than zero, such that the snippet above ends up comparing two _Bools with different representations: _Bool[3] == _Bool[1].



I can't find much in the C standard about what the result of such a comparison is. Since _Bool is an integer type, I'd expect the rules for integers to apply, such that the equality comparison only returns true if the representations are equal, which is not the case here.



Since on my platform this program returns 0, it would appear that this rule is not applying here.



Why does this code behave like this ? (i.e. what am I missing? Which representations of _Bool are trap representations and which ones aren't? How many representations can represent true and false ? What role do padding bits play into this? etc. )



What can portable C programs assume about the representation of _Bool ?










share|improve this question




















  • 1





    A union will cause both c and b to occupy the same area of memory, assigning 3 to char means the value of b will also be 3, true if NOT false, so it should equate to -1.

    – SPlatten
    Dec 30 '18 at 17:27






  • 1





    I could be wrong, but your program smells like undefined behavior to me

    – Basile Starynkevitch
    Dec 30 '18 at 17:32






  • 1





    In order to avoid integer overflows, any assignments to a _Bool that are not 0 (false) are stored as 1 (true). That being said, your code seems to be creating it's own confusion here, since the code says if (true) return false, and then you question why you are getting a false response. However, I believe that @interjay actually has the correct answer, and you have an undefined behavior due to your union causing the _Bool to return the lowest bit. This seems to imply that the Union is only useful to identify if a value is odd or even.

    – Claies
    Dec 30 '18 at 17:39








  • 1





    No, equality comparisons return true if the values are equal. Representations are irrelevant for this purpose

    – n.m.
    Dec 30 '18 at 17:44






  • 1





    My answer to Is it safe to memset bool to 0? has a lot of potentially useful background information.

    – Shafik Yaghmour
    Dec 30 '18 at 18:03
















6















This program returns 0 in my machine:



#include <stdbool.h>

union U {
_Bool b;
char c;
};

int main(void) {
union U u;
u.c = 3;
_Bool b = u.b;
if (b == true) {
return 0;
} else {
return 1;
}
}


AFAICT, _Bool is an integer type that can at least store 0 and 1, and true is the integral constant 1. On my machine, _Bool has a sizeof(_Bool) == 1, and CHAR_BITS == 8, which means that _Bool has 256 representations.



I can't find much in the C standard about the trap representations of _Bool, and I can't find whether creating a _Bool with a representation different from 0 or 1 (on implementations that support more than two representations) is ok, and if it is ok, whether those representations denote true or false.



What I can find in the standard is what happens when a _Bool is compared with an integer, the integer is converted to the 0 representation if it has value 0, and to the 1 representation if it has a value different than zero, such that the snippet above ends up comparing two _Bools with different representations: _Bool[3] == _Bool[1].



I can't find much in the C standard about what the result of such a comparison is. Since _Bool is an integer type, I'd expect the rules for integers to apply, such that the equality comparison only returns true if the representations are equal, which is not the case here.



Since on my platform this program returns 0, it would appear that this rule is not applying here.



Why does this code behave like this ? (i.e. what am I missing? Which representations of _Bool are trap representations and which ones aren't? How many representations can represent true and false ? What role do padding bits play into this? etc. )



What can portable C programs assume about the representation of _Bool ?










share|improve this question




















  • 1





    A union will cause both c and b to occupy the same area of memory, assigning 3 to char means the value of b will also be 3, true if NOT false, so it should equate to -1.

    – SPlatten
    Dec 30 '18 at 17:27






  • 1





    I could be wrong, but your program smells like undefined behavior to me

    – Basile Starynkevitch
    Dec 30 '18 at 17:32






  • 1





    In order to avoid integer overflows, any assignments to a _Bool that are not 0 (false) are stored as 1 (true). That being said, your code seems to be creating it's own confusion here, since the code says if (true) return false, and then you question why you are getting a false response. However, I believe that @interjay actually has the correct answer, and you have an undefined behavior due to your union causing the _Bool to return the lowest bit. This seems to imply that the Union is only useful to identify if a value is odd or even.

    – Claies
    Dec 30 '18 at 17:39








  • 1





    No, equality comparisons return true if the values are equal. Representations are irrelevant for this purpose

    – n.m.
    Dec 30 '18 at 17:44






  • 1





    My answer to Is it safe to memset bool to 0? has a lot of potentially useful background information.

    – Shafik Yaghmour
    Dec 30 '18 at 18:03














6












6








6


1






This program returns 0 in my machine:



#include <stdbool.h>

union U {
_Bool b;
char c;
};

int main(void) {
union U u;
u.c = 3;
_Bool b = u.b;
if (b == true) {
return 0;
} else {
return 1;
}
}


AFAICT, _Bool is an integer type that can at least store 0 and 1, and true is the integral constant 1. On my machine, _Bool has a sizeof(_Bool) == 1, and CHAR_BITS == 8, which means that _Bool has 256 representations.



I can't find much in the C standard about the trap representations of _Bool, and I can't find whether creating a _Bool with a representation different from 0 or 1 (on implementations that support more than two representations) is ok, and if it is ok, whether those representations denote true or false.



What I can find in the standard is what happens when a _Bool is compared with an integer, the integer is converted to the 0 representation if it has value 0, and to the 1 representation if it has a value different than zero, such that the snippet above ends up comparing two _Bools with different representations: _Bool[3] == _Bool[1].



I can't find much in the C standard about what the result of such a comparison is. Since _Bool is an integer type, I'd expect the rules for integers to apply, such that the equality comparison only returns true if the representations are equal, which is not the case here.



Since on my platform this program returns 0, it would appear that this rule is not applying here.



Why does this code behave like this ? (i.e. what am I missing? Which representations of _Bool are trap representations and which ones aren't? How many representations can represent true and false ? What role do padding bits play into this? etc. )



What can portable C programs assume about the representation of _Bool ?










share|improve this question
















This program returns 0 in my machine:



#include <stdbool.h>

union U {
_Bool b;
char c;
};

int main(void) {
union U u;
u.c = 3;
_Bool b = u.b;
if (b == true) {
return 0;
} else {
return 1;
}
}


AFAICT, _Bool is an integer type that can at least store 0 and 1, and true is the integral constant 1. On my machine, _Bool has a sizeof(_Bool) == 1, and CHAR_BITS == 8, which means that _Bool has 256 representations.



I can't find much in the C standard about the trap representations of _Bool, and I can't find whether creating a _Bool with a representation different from 0 or 1 (on implementations that support more than two representations) is ok, and if it is ok, whether those representations denote true or false.



What I can find in the standard is what happens when a _Bool is compared with an integer, the integer is converted to the 0 representation if it has value 0, and to the 1 representation if it has a value different than zero, such that the snippet above ends up comparing two _Bools with different representations: _Bool[3] == _Bool[1].



I can't find much in the C standard about what the result of such a comparison is. Since _Bool is an integer type, I'd expect the rules for integers to apply, such that the equality comparison only returns true if the representations are equal, which is not the case here.



Since on my platform this program returns 0, it would appear that this rule is not applying here.



Why does this code behave like this ? (i.e. what am I missing? Which representations of _Bool are trap representations and which ones aren't? How many representations can represent true and false ? What role do padding bits play into this? etc. )



What can portable C programs assume about the representation of _Bool ?







c boolean






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Dec 30 '18 at 17:25







gnzlbg

















asked Dec 30 '18 at 17:19









gnzlbggnzlbg

2,65723077




2,65723077








  • 1





    A union will cause both c and b to occupy the same area of memory, assigning 3 to char means the value of b will also be 3, true if NOT false, so it should equate to -1.

    – SPlatten
    Dec 30 '18 at 17:27






  • 1





    I could be wrong, but your program smells like undefined behavior to me

    – Basile Starynkevitch
    Dec 30 '18 at 17:32






  • 1





    In order to avoid integer overflows, any assignments to a _Bool that are not 0 (false) are stored as 1 (true). That being said, your code seems to be creating it's own confusion here, since the code says if (true) return false, and then you question why you are getting a false response. However, I believe that @interjay actually has the correct answer, and you have an undefined behavior due to your union causing the _Bool to return the lowest bit. This seems to imply that the Union is only useful to identify if a value is odd or even.

    – Claies
    Dec 30 '18 at 17:39








  • 1





    No, equality comparisons return true if the values are equal. Representations are irrelevant for this purpose

    – n.m.
    Dec 30 '18 at 17:44






  • 1





    My answer to Is it safe to memset bool to 0? has a lot of potentially useful background information.

    – Shafik Yaghmour
    Dec 30 '18 at 18:03














  • 1





    A union will cause both c and b to occupy the same area of memory, assigning 3 to char means the value of b will also be 3, true if NOT false, so it should equate to -1.

    – SPlatten
    Dec 30 '18 at 17:27






  • 1





    I could be wrong, but your program smells like undefined behavior to me

    – Basile Starynkevitch
    Dec 30 '18 at 17:32






  • 1





    In order to avoid integer overflows, any assignments to a _Bool that are not 0 (false) are stored as 1 (true). That being said, your code seems to be creating it's own confusion here, since the code says if (true) return false, and then you question why you are getting a false response. However, I believe that @interjay actually has the correct answer, and you have an undefined behavior due to your union causing the _Bool to return the lowest bit. This seems to imply that the Union is only useful to identify if a value is odd or even.

    – Claies
    Dec 30 '18 at 17:39








  • 1





    No, equality comparisons return true if the values are equal. Representations are irrelevant for this purpose

    – n.m.
    Dec 30 '18 at 17:44






  • 1





    My answer to Is it safe to memset bool to 0? has a lot of potentially useful background information.

    – Shafik Yaghmour
    Dec 30 '18 at 18:03








1




1





A union will cause both c and b to occupy the same area of memory, assigning 3 to char means the value of b will also be 3, true if NOT false, so it should equate to -1.

– SPlatten
Dec 30 '18 at 17:27





A union will cause both c and b to occupy the same area of memory, assigning 3 to char means the value of b will also be 3, true if NOT false, so it should equate to -1.

– SPlatten
Dec 30 '18 at 17:27




1




1





I could be wrong, but your program smells like undefined behavior to me

– Basile Starynkevitch
Dec 30 '18 at 17:32





I could be wrong, but your program smells like undefined behavior to me

– Basile Starynkevitch
Dec 30 '18 at 17:32




1




1





In order to avoid integer overflows, any assignments to a _Bool that are not 0 (false) are stored as 1 (true). That being said, your code seems to be creating it's own confusion here, since the code says if (true) return false, and then you question why you are getting a false response. However, I believe that @interjay actually has the correct answer, and you have an undefined behavior due to your union causing the _Bool to return the lowest bit. This seems to imply that the Union is only useful to identify if a value is odd or even.

– Claies
Dec 30 '18 at 17:39







In order to avoid integer overflows, any assignments to a _Bool that are not 0 (false) are stored as 1 (true). That being said, your code seems to be creating it's own confusion here, since the code says if (true) return false, and then you question why you are getting a false response. However, I believe that @interjay actually has the correct answer, and you have an undefined behavior due to your union causing the _Bool to return the lowest bit. This seems to imply that the Union is only useful to identify if a value is odd or even.

– Claies
Dec 30 '18 at 17:39






1




1





No, equality comparisons return true if the values are equal. Representations are irrelevant for this purpose

– n.m.
Dec 30 '18 at 17:44





No, equality comparisons return true if the values are equal. Representations are irrelevant for this purpose

– n.m.
Dec 30 '18 at 17:44




1




1





My answer to Is it safe to memset bool to 0? has a lot of potentially useful background information.

– Shafik Yaghmour
Dec 30 '18 at 18:03





My answer to Is it safe to memset bool to 0? has a lot of potentially useful background information.

– Shafik Yaghmour
Dec 30 '18 at 18:03












1 Answer
1






active

oldest

votes


















7














Footnote 122 in the C11 standard says:




While the number of bits in a _Bool object is at least CHAR_BIT, the width (number of sign and value bits) of a _Bool may be just 1 bit.




So on a compiler where _Bool has only one value bit, only one of the bits of the char will have effect when you read it from memory as a _Bool. The other bits are padding bits which are ignored.



When I test your code with GCC, the _Bool member gets a value of 1 when assigning an odd number to u.c and 0 when assigning an even number, suggesting that it only looks at the lowest bit.



Note that the above is true only for type-punning. If you instead convert (implicit or explicit cast) a char to a _Bool, the value will be 1 if the char was nonzero.






share|improve this answer


























  • The question is whether _Bool has only 1 bit of values in GCC. If it does, then all other representations are trap representations and the behavior of my example is undefined. If it does not, then _Bool has many "true" and "false" representations, instead of just 2.

    – gnzlbg
    Dec 31 '18 at 9:16






  • 2





    @gnzlbg It does have only 1 bit of value in GCC (you can check this by trying to create a bitfield with a _Bool b : 2 field, which fails). But the other representations are not necessarily trap representations: The unused bits are padding bits, which can have any value. So the 0 and 1 values have many possible representations. It's possible for the compiler to define certain combinations of padding bits as trap representations, but it doesn't look like GCC does so.

    – interjay
    Dec 31 '18 at 10:43











  • Thanks, that makes sense.

    – gnzlbg
    Dec 31 '18 at 12:23











  • FYI I just filled gcc.gnu.org/bugzilla/show_bug.cgi?id=88662 because it does look like GCC assumes that _Bool representations are either 0x0 or 0x1.

    – gnzlbg
    Jan 2 at 13:54











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

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active

oldest

votes









7














Footnote 122 in the C11 standard says:




While the number of bits in a _Bool object is at least CHAR_BIT, the width (number of sign and value bits) of a _Bool may be just 1 bit.




So on a compiler where _Bool has only one value bit, only one of the bits of the char will have effect when you read it from memory as a _Bool. The other bits are padding bits which are ignored.



When I test your code with GCC, the _Bool member gets a value of 1 when assigning an odd number to u.c and 0 when assigning an even number, suggesting that it only looks at the lowest bit.



Note that the above is true only for type-punning. If you instead convert (implicit or explicit cast) a char to a _Bool, the value will be 1 if the char was nonzero.






share|improve this answer


























  • The question is whether _Bool has only 1 bit of values in GCC. If it does, then all other representations are trap representations and the behavior of my example is undefined. If it does not, then _Bool has many "true" and "false" representations, instead of just 2.

    – gnzlbg
    Dec 31 '18 at 9:16






  • 2





    @gnzlbg It does have only 1 bit of value in GCC (you can check this by trying to create a bitfield with a _Bool b : 2 field, which fails). But the other representations are not necessarily trap representations: The unused bits are padding bits, which can have any value. So the 0 and 1 values have many possible representations. It's possible for the compiler to define certain combinations of padding bits as trap representations, but it doesn't look like GCC does so.

    – interjay
    Dec 31 '18 at 10:43











  • Thanks, that makes sense.

    – gnzlbg
    Dec 31 '18 at 12:23











  • FYI I just filled gcc.gnu.org/bugzilla/show_bug.cgi?id=88662 because it does look like GCC assumes that _Bool representations are either 0x0 or 0x1.

    – gnzlbg
    Jan 2 at 13:54
















7














Footnote 122 in the C11 standard says:




While the number of bits in a _Bool object is at least CHAR_BIT, the width (number of sign and value bits) of a _Bool may be just 1 bit.




So on a compiler where _Bool has only one value bit, only one of the bits of the char will have effect when you read it from memory as a _Bool. The other bits are padding bits which are ignored.



When I test your code with GCC, the _Bool member gets a value of 1 when assigning an odd number to u.c and 0 when assigning an even number, suggesting that it only looks at the lowest bit.



Note that the above is true only for type-punning. If you instead convert (implicit or explicit cast) a char to a _Bool, the value will be 1 if the char was nonzero.






share|improve this answer


























  • The question is whether _Bool has only 1 bit of values in GCC. If it does, then all other representations are trap representations and the behavior of my example is undefined. If it does not, then _Bool has many "true" and "false" representations, instead of just 2.

    – gnzlbg
    Dec 31 '18 at 9:16






  • 2





    @gnzlbg It does have only 1 bit of value in GCC (you can check this by trying to create a bitfield with a _Bool b : 2 field, which fails). But the other representations are not necessarily trap representations: The unused bits are padding bits, which can have any value. So the 0 and 1 values have many possible representations. It's possible for the compiler to define certain combinations of padding bits as trap representations, but it doesn't look like GCC does so.

    – interjay
    Dec 31 '18 at 10:43











  • Thanks, that makes sense.

    – gnzlbg
    Dec 31 '18 at 12:23











  • FYI I just filled gcc.gnu.org/bugzilla/show_bug.cgi?id=88662 because it does look like GCC assumes that _Bool representations are either 0x0 or 0x1.

    – gnzlbg
    Jan 2 at 13:54














7












7








7







Footnote 122 in the C11 standard says:




While the number of bits in a _Bool object is at least CHAR_BIT, the width (number of sign and value bits) of a _Bool may be just 1 bit.




So on a compiler where _Bool has only one value bit, only one of the bits of the char will have effect when you read it from memory as a _Bool. The other bits are padding bits which are ignored.



When I test your code with GCC, the _Bool member gets a value of 1 when assigning an odd number to u.c and 0 when assigning an even number, suggesting that it only looks at the lowest bit.



Note that the above is true only for type-punning. If you instead convert (implicit or explicit cast) a char to a _Bool, the value will be 1 if the char was nonzero.






share|improve this answer















Footnote 122 in the C11 standard says:




While the number of bits in a _Bool object is at least CHAR_BIT, the width (number of sign and value bits) of a _Bool may be just 1 bit.




So on a compiler where _Bool has only one value bit, only one of the bits of the char will have effect when you read it from memory as a _Bool. The other bits are padding bits which are ignored.



When I test your code with GCC, the _Bool member gets a value of 1 when assigning an odd number to u.c and 0 when assigning an even number, suggesting that it only looks at the lowest bit.



Note that the above is true only for type-punning. If you instead convert (implicit or explicit cast) a char to a _Bool, the value will be 1 if the char was nonzero.







share|improve this answer














share|improve this answer



share|improve this answer








edited Dec 30 '18 at 17:49

























answered Dec 30 '18 at 17:39









interjayinterjay

84.2k16202211




84.2k16202211













  • The question is whether _Bool has only 1 bit of values in GCC. If it does, then all other representations are trap representations and the behavior of my example is undefined. If it does not, then _Bool has many "true" and "false" representations, instead of just 2.

    – gnzlbg
    Dec 31 '18 at 9:16






  • 2





    @gnzlbg It does have only 1 bit of value in GCC (you can check this by trying to create a bitfield with a _Bool b : 2 field, which fails). But the other representations are not necessarily trap representations: The unused bits are padding bits, which can have any value. So the 0 and 1 values have many possible representations. It's possible for the compiler to define certain combinations of padding bits as trap representations, but it doesn't look like GCC does so.

    – interjay
    Dec 31 '18 at 10:43











  • Thanks, that makes sense.

    – gnzlbg
    Dec 31 '18 at 12:23











  • FYI I just filled gcc.gnu.org/bugzilla/show_bug.cgi?id=88662 because it does look like GCC assumes that _Bool representations are either 0x0 or 0x1.

    – gnzlbg
    Jan 2 at 13:54



















  • The question is whether _Bool has only 1 bit of values in GCC. If it does, then all other representations are trap representations and the behavior of my example is undefined. If it does not, then _Bool has many "true" and "false" representations, instead of just 2.

    – gnzlbg
    Dec 31 '18 at 9:16






  • 2





    @gnzlbg It does have only 1 bit of value in GCC (you can check this by trying to create a bitfield with a _Bool b : 2 field, which fails). But the other representations are not necessarily trap representations: The unused bits are padding bits, which can have any value. So the 0 and 1 values have many possible representations. It's possible for the compiler to define certain combinations of padding bits as trap representations, but it doesn't look like GCC does so.

    – interjay
    Dec 31 '18 at 10:43











  • Thanks, that makes sense.

    – gnzlbg
    Dec 31 '18 at 12:23











  • FYI I just filled gcc.gnu.org/bugzilla/show_bug.cgi?id=88662 because it does look like GCC assumes that _Bool representations are either 0x0 or 0x1.

    – gnzlbg
    Jan 2 at 13:54

















The question is whether _Bool has only 1 bit of values in GCC. If it does, then all other representations are trap representations and the behavior of my example is undefined. If it does not, then _Bool has many "true" and "false" representations, instead of just 2.

– gnzlbg
Dec 31 '18 at 9:16





The question is whether _Bool has only 1 bit of values in GCC. If it does, then all other representations are trap representations and the behavior of my example is undefined. If it does not, then _Bool has many "true" and "false" representations, instead of just 2.

– gnzlbg
Dec 31 '18 at 9:16




2




2





@gnzlbg It does have only 1 bit of value in GCC (you can check this by trying to create a bitfield with a _Bool b : 2 field, which fails). But the other representations are not necessarily trap representations: The unused bits are padding bits, which can have any value. So the 0 and 1 values have many possible representations. It's possible for the compiler to define certain combinations of padding bits as trap representations, but it doesn't look like GCC does so.

– interjay
Dec 31 '18 at 10:43





@gnzlbg It does have only 1 bit of value in GCC (you can check this by trying to create a bitfield with a _Bool b : 2 field, which fails). But the other representations are not necessarily trap representations: The unused bits are padding bits, which can have any value. So the 0 and 1 values have many possible representations. It's possible for the compiler to define certain combinations of padding bits as trap representations, but it doesn't look like GCC does so.

– interjay
Dec 31 '18 at 10:43













Thanks, that makes sense.

– gnzlbg
Dec 31 '18 at 12:23





Thanks, that makes sense.

– gnzlbg
Dec 31 '18 at 12:23













FYI I just filled gcc.gnu.org/bugzilla/show_bug.cgi?id=88662 because it does look like GCC assumes that _Bool representations are either 0x0 or 0x1.

– gnzlbg
Jan 2 at 13:54





FYI I just filled gcc.gnu.org/bugzilla/show_bug.cgi?id=88662 because it does look like GCC assumes that _Bool representations are either 0x0 or 0x1.

– gnzlbg
Jan 2 at 13:54


















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