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From: Anubhav on 6 Aug 2010 19:18 int& f(){ static int x = 0; x++; return x; } int main(){ int x = 0; f() += 1; f() = f() + 1; cout << f(); } What should be the output of the code shown? VS gives 5 and gcc gives 6. Regards, Dabs. -- [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ]
From: Stefan van Kessel on 7 Aug 2010 12:09 The code you posted relies upon the order of evaluation within the statement "f() = f() + 1;". Either the first or the second invocation of f() has to come first. If the first comes first, you have x = 4 + 1; (x is now 5) If the second comes first you have essentially ++x = 3 + 1. (x is now 4) Since I'm pretty sure the order of evaluation here is undefined, the behavior also is undefined. If you've got a copy of the standard at hand, look at 1.9.15. On 8/7/2010 12:18 PM, Anubhav wrote: > int& f(){ > static int x = 0; > x++; > return x; > } > > int main(){ > int x = 0; > f() += 1; > f() = f() + 1; > > cout<< f(); > } > > What should be the output of the code shown? > > VS gives 5 and gcc gives 6. > > Regards, > Dabs. > -- [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ]
From: Bo Persson on 7 Aug 2010 12:06 Anubhav wrote: > int& f(){ > static int x = 0; > x++; > return x; > } > > int main(){ > int x = 0; > f() += 1; > f() = f() + 1; > > cout << f(); > } > > What should be the output of the code shown? > > VS gives 5 and gcc gives 6. > > Regards, > Dabs. We don't know. In the third statement, the f's could be called in any order. Either would be correct. Bo Persson -- [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ]
From: Johannes Schaub (litb) on 7 Aug 2010 12:07 Anubhav wrote: > int& f(){ > static int x = 0; > x++; > return x; > } > > int main(){ > int x = 0; > f() += 1; > f() = f() + 1; > > cout << f(); > } > > What should be the output of the code shown? > > VS gives 5 and gcc gives 6. > The output is unspecified, because the orderof value computation of the operands of op= is unspecified. Lvalue computation of the two f()'s are unsequenced. So how the "f()" value in "f() + 1" gets computed to can be either f() += 1; // after which f.x == 2 f() = f() + 1 // after which f.x == 5 or 4 cout << f() // after which f.x == 6 or 5 -- [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ]
From: Prasoon Saurav on 7 Aug 2010 12:32
On Aug 7, 3:18 pm, Anubhav <rkld...(a)gmail.com> wrote: > int& f(){ > static int x = 0; > x++; > return x; > > } > > int main(){ > int x = 0; > f() += 1; > f() = f() + 1; > > cout << f(); > > } > > What should be the output of the code shown? > > VS gives 5 and gcc gives 6. > > Regards, > Dabs. > As per my opinion your code invokes Undefined Behaviour because the order of evaluation of arguments of assignment operator is unspecified and both right hand side and left hand side f() refer to the same static variable. So indirectly you are trying to modify the variable more than once between two sequence points. Somebody please correct me if I am wrong. -- [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] |