C++Talk.NET C++ language newsgroups   Archives   FAQ   Search   Memberlist   Usergroups   Register   Profile   Log in to check your private messages   Log in  Signed mod unsigned Goto page 1, 2  Next          C++Talk.NET Forum Index -> C Language View previous topic :: View next topic   Author Message Johannes Bauer Guest Posted: Wed Jun 06, 2012 11:55 am    Post subject: Signed mod unsigned Hi group, today I ran into something I honestly did not expect. Consider this snippet: #include <stdio.h> int main() { int i1, i2; unsigned int u2; i1 = -1; i2 = 30; u2 = 30; printf("%d\n", i1 % i2); printf("%d\n", i1 % u2); return 0; } I expected to find the output "-1 / -1". Instead the result I got was "-1 / 15". Apparently when getting the remainder of a signed by a unsigned int, the signed is silently promoted to unsigned in twos complement and then the operation is performed, i.e. 0xffffffff % 30 == 15 I would have expected at least a warning since I find this highly unintuitive -- my expectation would be that the unsigned becomes a (truncated) signed and a warning. Instead gcc does not emit a warning (-Wall -Wextra) and silently performs above operation. Is this really in accordance with the C standard or is gcc doing something weird here? Best regards, Joe -- Quote: Wo hattest Du das Beben nochmal GENAU vorhergesagt? Zumindest nicht öffentlich! Ah, der neueste und bis heute genialste Streich unsere großen Kosmologen: Die Geheim-Vorhersage. - Karl Kaos über Rüdiger Thomas in dsa <hidbv3$om2$1 (AT) speranza (DOT) aioe.org> Back to top James Kuyper Guest Posted: Wed Jun 06, 2012 12:41 pm    Post subject: Re: Signed mod unsigned On 06/06/2012 09:55 AM, Johannes Bauer wrote: Quote: Hi group, today I ran into something I honestly did not expect. Consider this snippet: #include <stdio.h int main() { int i1, i2; unsigned int u2; i1 = -1; i2 = 30; u2 = 30; printf("%d\n", i1 % i2); printf("%d\n", i1 % u2); return 0; } I expected to find the output "-1 / -1". Instead the result I got was "-1 / 15". Apparently when getting the remainder of a signed by a unsigned int, the signed is silently promoted to unsigned in twos complement and then the operation is performed, i.e. 0xffffffff % 30 == 15 I would have expected at least a warning since I find this highly unintuitive -- my expectation would be that the unsigned becomes a (truncated) signed and a warning. Instead gcc does not emit a warning (-Wall -Wextra) and silently performs above operation. Is this really in accordance with the C standard or is gcc doing something weird here? In most binary expressions (modulus expressions are no exception - 6.5.5p3), the operands are subject to what is called the usual arithmetic conversions (6.3.1. before the operation is performed. Those conversions include the integer promotions, which leave both operands unchanged in this case. The relevant rule is that when one promoted type is unsigned, and the other is signed, if the unsigned type has an integer conversion rank greater than or equal to that of the signed type, then the operand of the signed type is converted to the unsigned type. "unsigned int" is required to have the same integer conversion rank as "int" (6.3.1.1p1); So the behavior you describe is not only allowed, but required, by the C standard. There's some minor details about your description that should be modified: this is not an example of what the C standard calls a promotion, and it has nothing to do with 2's complement. The C standard also allows 1's complement and sign-magnitude representations (6.2.6.2p2), and the conversion from a signed value to an unsigned type follows the same rules regardless of the representation used for the signed type. A value 1 greater than the maximum value representable in the unsigned type is either repeatedly added to, or repeatedly subtract from, the value until the result is representable in the unsigned type (6.3.1.3p2). The machine instructions needed to obtain this result can be quite different, depending upon which representation is used for the signed type. However, the result of the conversion should be exactly the same for all three representations, it depends only upon the initial value represented and maximum representable value of the unsigned type. Back to top gwowen Guest Posted: Wed Jun 06, 2012 2:53 pm    Post subject: Re: Signed mod unsigned On Jun 6, 2:55 pm, Johannes Bauer <dfnsonfsdu...@gmx.de> wrote: Quote: Is this really in accordance with the C standard or is gcc doing something weird here? James has given a complete technical answer, so I won't add to that. What I will say is that there's no universally accepted definition of what remainder/modulus means, so whatever intuition tells you[0], someone is going to consider it to be wrong. When it comes to remainders involving negatives, intuition is worthless. [0] My intuition, for example, tells me that -1 % 30 equals 29. My intuition is worthless too. Back to top James Kuyper Guest Posted: Wed Jun 06, 2012 8:29 pm    Post subject: Re: Signed mod unsigned On 06/06/2012 06:15 PM, Edward Rutherford wrote: Quote: Johannes Bauer wrote: Hi group, today I ran into something I honestly did not expect. Consider this snippet: #include <stdio.h int main() { int i1, i2; unsigned int u2; i1 = -1; i2 = 30; u2 = 30; printf("%d\n", i1 % i2); printf("%d\n", i1 % u2); return 0; } I expected to find the output "-1 / -1". Instead the result I got was "-1 / 15". Apparently when getting the remainder of a signed by a unsigned int, the signed is silently promoted to unsigned in twos complement and then the operation is performed, i.e. 0xffffffff % 30 == 15 I would have expected at least a warning since I find this highly unintuitive -- my expectation would be that the unsigned becomes a (truncated) signed and a warning. Instead gcc does not emit a warning (-Wall -Wextra) and silently performs above operation. Is this really in accordance with the C standard or is gcc doing something weird here? This looks like a compiler bug. I would report it to a Gcc mailing list. I explained in my response why I believe that this behavior is not merely allowed, but mandated, by the C standard. Do you disagree? If so, on what grounds? Back to top Edward Rutherford Guest Posted: Wed Jun 06, 2012 10:15 pm    Post subject: Re: Signed mod unsigned Johannes Bauer wrote: Quote: Hi group, today I ran into something I honestly did not expect. Consider this snippet: #include <stdio.h int main() { int i1, i2; unsigned int u2; i1 = -1; i2 = 30; u2 = 30; printf("%d\n", i1 % i2); printf("%d\n", i1 % u2); return 0; } I expected to find the output "-1 / -1". Instead the result I got was "-1 / 15". Apparently when getting the remainder of a signed by a unsigned int, the signed is silently promoted to unsigned in twos complement and then the operation is performed, i.e. 0xffffffff % 30 == 15 I would have expected at least a warning since I find this highly unintuitive -- my expectation would be that the unsigned becomes a (truncated) signed and a warning. Instead gcc does not emit a warning (-Wall -Wextra) and silently performs above operation. Is this really in accordance with the C standard or is gcc doing something weird here? This looks like a compiler bug. I would report it to a Gcc mailing list. Back to top Johannes Bauer Guest Posted: Thu Jun 07, 2012 8:12 am    Post subject: Re: Signed mod unsigned On 06.06.2012 16:41, James Kuyper wrote: [...] Quote: So the behavior you describe is not only allowed, but required, by the C standard. James, thank you very much for the verbose and detailed explanation. In 15 years of C programming, I did not run into that caveat :-) Best regards, Johannes -- Quote: Wo hattest Du das Beben nochmal GENAU vorhergesagt? Zumindest nicht öffentlich! Ah, der neueste und bis heute genialste Streich unsere großen Kosmologen: Die Geheim-Vorhersage. - Karl Kaos über Rüdiger Thomas in dsa <hidbv3$om2$1 (AT) speranza (DOT) aioe.org> Back to top Johannes Bauer Guest Posted: Thu Jun 07, 2012 8:15 am    Post subject: Re: Signed mod unsigned On 06.06.2012 16:53, gwowen wrote: Quote: When it comes to remainders involving negatives, intuition is worthless. [0] My intuition, for example, tells me that -1 % 30 equals 29. My intuition is worthless too. Well, it's always nice when intuition at least overlaps with the reality of compilers or interpreters :-) BTW, -1 % 30 == 29 in Python. I did know that C did not give up the sign, but I'd have expected for example -100 % 30 == -10 (it is 20 in Python). Anyways, it does so for signed/signed types, so I'm going to use these. Best regards, Joe -- Quote: Wo hattest Du das Beben nochmal GENAU vorhergesagt? Zumindest nicht öffentlich! Ah, der neueste und bis heute genialste Streich unsere großen Kosmologen: Die Geheim-Vorhersage. - Karl Kaos über Rüdiger Thomas in dsa <hidbv3$om2$1 (AT) speranza (DOT) aioe.org> Back to top James Kuyper Guest Posted: Thu Jun 07, 2012 8:44 am    Post subject: Re: Signed mod unsigned On 06/07/2012 06:12 AM, Johannes Bauer wrote: Quote: On 06.06.2012 16:41, James Kuyper wrote: [...] So the behavior you describe is not only allowed, but required, by the C standard. James, thank you very much for the verbose and detailed explanation. In 15 years of C programming, I did not run into that caveat This issue comes up in almost every binary operation involving negative values and unsigned integers of a type with equal or greater integer conversion rank; if you've not noticed it before in 15 years of C programming, then you've probably been deliberately avoiding mixing negative values with unsigned types - which is generally a good idea. -- James Kuyper Back to top BartC Guest Posted: Thu Jun 07, 2012 9:29 am    Post subject: Re: Signed mod unsigned Johannes Bauer" <dfnsonfsduifb (AT) gmx (DOT) de> wrote in message news:jqpv0d$beq$1 (AT) news (DOT) albasani.net... Quote: On 06.06.2012 16:53, gwowen wrote: [0] My intuition, for example, tells me that -1 % 30 equals 29. My intuition is worthless too. Well, it's always nice when intuition at least overlaps with the reality of compilers or interpreters :-) BTW, -1 % 30 == 29 in Python. I did know that C did not give up the sign, but I'd have expected for example -100 % 30 == -10 (it is 20 in Python). Anyways, it does so for signed/signed types, so I'm going to use these. So -1%30 is -1 in C and 29 in Python. While -1%30u in C is 15. -1, 15 and 29; any other possible results for the same expression? -- Bartc Back to top Johannes Bauer Guest Posted: Thu Jun 07, 2012 9:31 am    Post subject: Re: Signed mod unsigned On 07.06.2012 12:44, James Kuyper wrote: Quote: if you've not noticed it before in 15 years of C programming, then you've probably been deliberately avoiding mixing negative values with unsigned types - which is generally a good idea. I try to do arithmetic, indexing and counting (i.e. loop variables) with signed types and use unsigned almost always only for bit operations (or where the well-defined overflow is needed). Has worked well so far :-) Best regards, Joe -- Quote: Wo hattest Du das Beben nochmal GENAU vorhergesagt? Zumindest nicht öffentlich! Ah, der neueste und bis heute genialste Streich unsere großen Kosmologen: Die Geheim-Vorhersage. - Karl Kaos über Rüdiger Thomas in dsa <hidbv3$om2$1 (AT) speranza (DOT) aioe.org> Back to top Johannes Bauer Guest Posted: Thu Jun 07, 2012 9:43 am    Post subject: Re: Signed mod unsigned On 07.06.2012 13:29, BartC wrote: Quote: BTW, -1 % 30 == 29 in Python. I did know that C did not give up the sign, but I'd have expected for example -100 % 30 == -10 (it is 20 in Python). Anyways, it does so for signed/signed types, so I'm going to use these. So -1%30 is -1 in C and 29 in Python. While -1%30u in C is 15. -1, 15 and 29; any other possible results for the same expression? If I understand James' post correct, in the expression -1 % 30u the "-1" is converted to UNSIGNED_MAX (which happens to be (2^32)-1 on my system). Since int IIRC is only required to have >= 16 bits, one could image a n-bit system in which the expression would evaluate to something different. In particular, if bits % 4 == 0, it evaluates to 15. For bits % 4 result 0 15 1 1 2 3 3 7 So I think the expression can be either -1, 15, 29, 1, 3 or 7 depending on the language and platform :-) Best regards, Johannes -- Quote: Wo hattest Du das Beben nochmal GENAU vorhergesagt? Zumindest nicht öffentlich! Ah, der neueste und bis heute genialste Streich unsere großen Kosmologen: Die Geheim-Vorhersage. - Karl Kaos über Rüdiger Thomas in dsa <hidbv3$om2$1 (AT) speranza (DOT) aioe.org> Back to top Eric Sosman Guest Posted: Thu Jun 07, 2012 9:46 am    Post subject: Re: Signed mod unsigned On 6/7/2012 7:29 AM, BartC wrote: Quote: Johannes Bauer" <dfnsonfsduifb (AT) gmx (DOT) de> wrote in message news:jqpv0d$beq$1 (AT) news (DOT) albasani.net... On 06.06.2012 16:53, gwowen wrote: [0] My intuition, for example, tells me that -1 % 30 equals 29. My intuition is worthless too. Well, it's always nice when intuition at least overlaps with the reality of compilers or interpreters :-) BTW, -1 % 30 == 29 in Python. I did know that C did not give up the sign, but I'd have expected for example -100 % 30 == -10 (it is 20 in Python). Anyways, it does so for signed/signed types, so I'm going to use these. So -1%30 is -1 in C and 29 in Python. While -1%30u in C is 15. -1%30u in C *can be* 15. Quote: -1, 15 and 29; any other possible results for the same expression? In C, the possible results are 1, 3, 7, 15. -- Eric Sosman esosman@ieee-dot-org.invalid Back to top Ben Bacarisse Guest Posted: Thu Jun 07, 2012 5:58 pm    Post subject: Re: Signed mod unsigned Eric Sosman <esosman@ieee-dot-org.invalid> writes: Quote: On 6/7/2012 7:29 AM, BartC wrote: Johannes Bauer" <dfnsonfsduifb (AT) gmx (DOT) de> wrote in message news:jqpv0d$beq$1 (AT) news (DOT) albasani.net... On 06.06.2012 16:53, gwowen wrote: [0] My intuition, for example, tells me that -1 % 30 equals 29. My intuition is worthless too. Well, it's always nice when intuition at least overlaps with the reality of compilers or interpreters :-) BTW, -1 % 30 == 29 in Python. I did know that C did not give up the sign, but I'd have expected for example -100 % 30 == -10 (it is 20 in Python). Anyways, it does so for signed/signed types, so I'm going to use these. So -1%30 is -1 in C and 29 in Python. While -1%30u in C is 15. -1%30u in C *can be* 15. -1, 15 and 29; any other possible results for the same expression? In C, the possible results are 1, 3, 7, 15. I think -1 is also possible on those awkward systems where UINT_MAX == INT_MAX. -- Ben. Back to top James Kuyper Guest Posted: Thu Jun 07, 2012 7:08 pm    Post subject: Re: Signed mod unsigned On 06/07/2012 04:44 PM, Edward Rutherford wrote: Quote: James Kuyper wrote: On 06/06/2012 06:15 PM, Edward Rutherford wrote: Johannes Bauer wrote: Hi group, today I ran into something I honestly did not expect. Consider this snippet: #include <stdio.h int main() { int i1, i2; unsigned int u2; i1 = -1; i2 = 30; u2 = 30; printf("%d\n", i1 % i2); printf("%d\n", i1 % u2); return 0; } I expected to find the output "-1 / -1". Instead the result I got was "-1 / 15". Apparently when getting the remainder of a signed by a unsigned int, the signed is silently promoted to unsigned in twos complement and then the operation is performed, i.e. 0xffffffff % 30 == 15 I would have expected at least a warning since I find this highly unintuitive -- my expectation would be that the unsigned becomes a (truncated) signed and a warning. Instead gcc does not emit a warning (-Wall -Wextra) and silently performs above operation. Is this really in accordance with the C standard or is gcc doing something weird here? This looks like a compiler bug. I would report it to a Gcc mailing list. I explained in my response why I believe that this behavior is not merely allowed, but mandated, by the C standard. Do you disagree? If so, on what grounds? This outcome just seems counterintuitive to me. Even if it is permitted by the C standard, this option is kinda retarded and I'd say the compiler has poor quality of implementation. It's not merely permitted; it's not an option; it's required. The phrase "Quality of implementation" is normally considered to apply only to issues where the standard gives a conforming implementation freedom to choose how something is done. The only freedom that the implementation has that's relevant to this program is in number of value bits in an unsigned int. You're not suggesting, I hope, that choosing 32 for that number gives the implementation a low QoI? It's a pretty popular choice. Back to top Edward Rutherford Guest Posted: Thu Jun 07, 2012 8:44 pm    Post subject: Re: Signed mod unsigned James Kuyper wrote: Quote: On 06/06/2012 06:15 PM, Edward Rutherford wrote: Johannes Bauer wrote: Hi group, today I ran into something I honestly did not expect. Consider this snippet: #include <stdio.h int main() { int i1, i2; unsigned int u2; i1 = -1; i2 = 30; u2 = 30; printf("%d\n", i1 % i2); printf("%d\n", i1 % u2); return 0; } I expected to find the output "-1 / -1". Instead the result I got was "-1 / 15". Apparently when getting the remainder of a signed by a unsigned int, the signed is silently promoted to unsigned in twos complement and then the operation is performed, i.e. 0xffffffff % 30 == 15 I would have expected at least a warning since I find this highly unintuitive -- my expectation would be that the unsigned becomes a (truncated) signed and a warning. Instead gcc does not emit a warning (-Wall -Wextra) and silently performs above operation. Is this really in accordance with the C standard or is gcc doing something weird here? This looks like a compiler bug. I would report it to a Gcc mailing list. I explained in my response why I believe that this behavior is not merely allowed, but mandated, by the C standard. Do you disagree? If so, on what grounds? This outcome just seems counterintuitive to me. Even if it is permitted by the C standard, this option is kinda retarded and I'd say the compiler has poor quality of implementation. Best regards, E.P.R. Back to top        C++Talk.NET Forum Index -> C Language All times are GMT Goto page 1, 2  Next Page 1 of 2 Powered by phpBB © 2001, 2006 phpBB Group