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Re: about increment- SHYAN PLEASE SEE THIS

ankushj10 Posted: Wed Jan 05, 2005 5:30 pm

Joined: 05 Jan 2005

Posts: 1

Re: about increment- SHYAN PLEASE SEE THIS

hi,
i think there is an order of evaluation, called "ASSOCIATIVITY(of operators- see kernighan and ritchie)". In other words it is order or scanning (left-right or right -left) and order of execution(precedence).

i still have litle knowledge, as i found it confusing.

TRY TO JUSTIFY THE OUTPUT OF THE PROGRAM BELOW:-

void temp(){ int x; x=5; printf("%d,%d,%d,%d,%d
",++x,++x,--x,x,x);}

// OUTPUT 6,5,4,5,5

bye
ankush.

Shyan Lam <sflam@...> wrote:
Reply embedded...> -----Original Message-----> From: ashwin mittal [mailto:ashwin_mittal26@...]> Sent: Sunday, December 26, 2004 10:50 PM> To: Programmers-Town@yahoogroups.com> Subject: Re: (PT) Re: about increment> > --- In Programmers-Town@yahoogroups.com, srikanth> > poojary> > <srikipoojary@y...> wrote:> > > main()> > > {> > > int i=5;> > > printf("%d%d%d%d%d",i++,i--,++i,--i,i);> > > }> > >> > >> > > what is the o/p and why?> >> Hi all,> > The Printf expression is compile from Right to left...You`re totally wrong here. The language does not specify any particularargument passing order. > So starting from right,> i,>
--i,> ++i,> i--,> i++> the sequence goes like this....Function arguments are not evaluated in any particular order either. Thereis no "sequence" for evaluation of function arguments.> So> i = 5;> --i = 4;> ++i = 5;> i-- = 5;> i++ = 4Again, wrong. There is no sequence point when evaluating the argument. Soafter the first `i++`, the value of `i` is undetermined until the nextsequence point is reached (that is at the time just before the functioncall).Because of the undefined behavior, the entire statement is undefined. If you do not know what a sequence point is and what classified as undefinedbehavior, you don`t know enough about C. Shyan

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