C++Talk.NET C++ language newsgroups   Archives   FAQ   Search   Memberlist   Usergroups   Register   Profile   Log in to check your private messages   Log in  the usual conversions          C++Talk.NET Forum Index -> C++ Language (Moderated) View previous topic :: View next topic   Author Message Marco Manfredini Guest Posted: Tue Jul 27, 2004 4:32 pm    Post subject: the usual conversions I found this issue here with gcc 3.4.1 and I'm unsure about it: #include <iostream> using namespace std; int f0(int x) { int k=1000000; return (x*1000000)/1000000; } int f1(int x) { const int k=1000000; return (x*k)/k; } int f2(int x) { int k=1000000; return (x*k)/k; } int main() { cout << f0(10000) << endl; cout << f1(10000) << endl; cout << f2(10000) << endl; } The output of this is when compiled (even without optimizations) is: 10000 10000 1410 GCC uses the equation (K*x)/K=x to fold the expressions with *constant* K, giving different output, because the overflow does not occur[1]. Did I overlook anything that could justify this behaviour? Or is it just dead wrong, as I think, because it omits the "ususal conversions"? Marco [1] No problem here, if x was declared as a small type, where x * 1000000 could not possibly overflow. [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top tom_usenet Guest Posted: Wed Jul 28, 2004 2:26 am    Post subject: Re: the usual conversions On 27 Jul 2004 12:32:56 -0400, Marco Manfredini <nospam (AT) phoyd (DOT) net> wrote: Quote: I found this issue here with gcc 3.4.1 and I'm unsure about it: #include using namespace std; int f0(int x) { int k=1000000; return (x*1000000)/1000000; } int f1(int x) { const int k=1000000; return (x*k)/k; } int f2(int x) { int k=1000000; return (x*k)/k; } int main() { cout << f0(10000) << endl; cout << f1(10000) << endl; cout << f2(10000) << endl; } The output of this is when compiled (even without optimizations) is: 10000 10000 1410 That seems correct (as would any other output). Quote: GCC uses the equation (K*x)/K=x to fold the expressions with *constant* K, giving different output, because the overflow does not occur[1]. Did I overlook anything that could justify this behaviour? Or is it just dead wrong, as I think, because it omits the "ususal conversions"? All three functions have signed int overflow, therefore all three have undefined behaviour and can validly output anything at all or just crash your computer. Tom [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Hyman Rosen Guest Posted: Wed Jul 28, 2004 2:54 am    Post subject: Re: the usual conversions Marco Manfredini wrote: Quote: GCC uses the equation (K*x)/K=x to fold the expressions with *constant* K, giving different output, because the overflow does not occur[1]. Overflow on signed arithmetic is undefined behavior. [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Display posts from previous: All Posts1 Day7 Days2 Weeks1 Month3 Months6 Months1 Year Oldest FirstNewest First         C++Talk.NET Forum Index -> C++ Language (Moderated) All times are GMT Page 1 of 1   Jump to: Select a forum C/C++----------------C LanguageC Language (Moderated)C++ Language (Moderated)C++ language (comp.lang.c++)C++ language, library and standards Non-English C/C++ Newsgroups----------------C++ (German)C++ (French)  You cannot post new topics in this forum You cannot reply to topics in this forum You cannot edit your posts in this forum You cannot delete your posts in this forum You cannot vote in polls in this forum Powered by phpBB © 2001, 2006 phpBB Group SEO toolkit © 2004-2006 webmedic.