plz explain this TC++ output

[prannoy]

this is a code which i met by. The question is to find the output. but the one i got by myself had little resemblence with that given by the computer. i asked my teacher my friends. but none of them could give a sensible reason. i reckon there could be a wierd rearrangement or precedence rules in Turbo C++ which i am not aware of.

plz give it a try. and dont forget to post some replies.

the code is:
void main()
{ int x[]= {10,20,30,40};
int *p,*t,**q;
p=x;
t=x+1;
q=&t;
cout <<*p<<'\t'<<**q<<'\t'<<*t++;
}





so got 10 20 20 eh?

but sorry my dear friend, the computer gives me
10 30 20

thanx for posting a reply!!

[Savage]

this is a code which i met by. The question is to find the output. but the one i got by myself had little resemblence with that given by the computer. i asked my teacher my friends. but none of them could give a sensible reason. i reckon there could be a wierd rearrangement or precedence rules in Turbo C++ which i am not aware of.

plz give it a try. and dont forget to post some replies.

the code is:
void main()
{ int x[]= {10,20,30,40};
int *p,*t,**q;
p=x;
t=x+1;
q=&t;
cout <<*p<<'\t'<<**q<<'\t'<<*t++;
}





so got 10 20 20 eh?

but sorry my dear friend, the computer gives me
10 30 20

thanx for posting a reply!!

You havent assigned pointers well.You should use:

*p=x;
*t=x+1;

[JosAH]

You too hit the 'undefined behaviour' behaviour when you alter the value of an
alterable lvalue more than once or both dereference and alter it before a
sequence point has been reached.

In the middle of an expression there are no sequence points and your code alters
the value of 't' while dereferencing it for lvalue 't'

C and C++ are not Java: they can evaluate any part of an expression in any
order they want as long as it follows the semantics of a defined abstract
machine (as per the ANSI/ISO standard). Java's defined behaviour is that
it has no undefined behaviour and always evaluates an expression from left
to right respecting operator precedence.

Your code simply violated that C and C++ constraint so nothing more can be
said about any behaviour on any platform whatsoever.

kind regards,

Jos

[prannoy]

You havent assigned pointers well.You should use:

*p=x;
*t=x+1;

actually i hav tested the code on tc++ compiler. and i got 10 30 20 without errors. anyway thanx for your reply.

[prannoy]

You too hit the 'undefined behaviour' behaviour when you alter the value of an
alterable lvalue more than once or both dereference and alter it before a
sequence point has been reached.

In the middle of an expression there are no sequence points and your code alters
the value of 't' while dereferencing it for lvalue 't'

C and C++ are not Java: they can evaluate any part of an expression in any
order they want as long as it follows the semantics of a defined abstract
machine (as per the ANSI/ISO standard). Java's defined behaviour is that
it has no undefined behaviour and always evaluates an expression from left
to right respecting operator precedence.

Your code simply violated that C and C++ constraint so nothing more can be
said about any behaviour on any platform whatsoever.

kind regards,

Jos

THANK YOU JOS!!!!!!!!!
i think ur reply most matches my problem. i really thank you for answering the 'unanswerable' question of mine. kudos to you!!

regards,

Prannoy