Need some explanation

[rahul8143]

hello,
1) First how following program get executed i mean how output is
printed and also why following program gives different output in Turbo
C++ compiler and Visual c++ 6 compiler?
void main()
{
int val=5;
printf("%d %d %d %d",val,--val,++val,val--);
}
under turbo compiler its giving
4 4 5 5
and under visual c++ its
5 5 6 5

2) How to evaluate following statement
int val =5;
val =- --val - val-- - --val;

[Suman]

ra*******@gmail .com wrote:

hello,
1) First how following program get executed i mean how output is
printed and also why following program gives different output in Turbo
C++ compiler and Visual c++ 6 compiler? #include <stdio.h> void main() is not a portable standard definition for main, use:
int main(/*int argc, char **argv*/) {
int val=5;
printf("%d %d %d %d",val,--val,++val,val--); /*falling off main, is a rather recent addition C99 onwards, AFAIK,
hence*/
return 0; }
under turbo compiler its giving
4 4 5 5
and under visual c++ its
5 5 6 5 The order in which the arguments to printf are evaluated is *not* the
same across different compiler/system pairs. 2) How to evaluate following statement
int val =5;
val =- --val - val-- - --val;

You might like to read the FAQ, and some stuff on side-effects and
sequence points.

[Suman]

ra*******@gmail .com wrote:

hello,
1) First how following program get executed i mean how output is
printed and also why following program gives different output in Turbo
C++ compiler and Visual c++ 6 compiler?
void main()
{
int val=5;
printf("%d %d %d %d",val,--val,++val,val--);
}
under turbo compiler its giving
4 4 5 5
and under visual c++ its
5 5 6 5

2) How to evaluate following statement
int val =5;
val =- --val - val-- - --val;

Goto http://www.eskimo.com/~scs/C-faq/s3.html.

[akarl]

ra*******@gmail .com wrote:

hello,
1) First how following program get executed i mean how output is
printed and also why following program gives different output in Turbo
C++ compiler and Visual c++ 6 compiler?
void main()
{
int val=5;
printf("%d %d %d %d",val,--val,++val,val--);
}
under turbo compiler its giving
4 4 5 5
and under visual c++ its
5 5 6 5

2) How to evaluate following statement
int val =5;
val =- --val - val-- - --val;

In the first case the behavior is undefined (evaluation order of
function parameters). The statement in (2) is very hard to read and
should (in practice) be rewritten into something more comprehensible.

Moreover:

- `void main()' is not a valid ANSI C signature for the main function.

- You have not included stdio.h

- Your coding style is inconsistent and the indentation is strange to
say the least.

[John Bode]

ra*******@gmail .com wrote:

hello,
1) First how following program get executed i mean how output is
printed and also why following program gives different output in Turbo
C++ compiler and Visual c++ 6 compiler?
It's because you are invoking undefined behavior in both cases. Read
the following: http://www.eskimo.com/~scs/C-faq/s3.html

The expression i++ evaluates to the current value of i; the side affect
(i incremented by 1) is applied sometime before the next sequence
point, but it need not be immediately after the expression is
evaluated.

Similarly for ++i; it evaluates to the current value of i + 1, but i is
not necessarily incremented immediately.

There is no requirement on when the side affects are applied other than
it must happen before the next sequence point.

For example, given the statement

i = j++ + ++k;

the following sequence is possible:

t1 <- j
t2 <- k + 1
i <- t1 + t2
j <- j + 1
k <- k + 1

Same thing applies to your printf() statement; there's no sequence
point between the autoincrement/decrement expressions, so the behavior
is undefined, meaning the compiler can do anything it wants.
void main()
You mean int main(void). Strictly speaking, "void main()" is an error.
{
int val=5;
printf("%d %d %d %d",val,--val,++val,val--);
}
under turbo compiler its giving
4 4 5 5
and under visual c++ its
5 5 6 5

2) How to evaluate following statement
int val =5;
val =- --val - val-- - --val;

[Christian Kandeler]

akarl wrote:

1) First how following program get executed i mean how output is
printed and also why following program gives different output in Turbo
C++ compiler and Visual c++ 6 compiler?
void main()
{
int val=5;
printf("%d %d %d %d",val,--val,++val,val--);
}
under turbo compiler its giving
4 4 5 5
and under visual c++ its
5 5 6 5

2) How to evaluate following statement
int val =5;
val =- --val - val-- - --val;

In the first case the behavior is undefined (evaluation order of
function parameters). The statement in (2) is very hard to read and
should (in practice) be rewritten into something more comprehensible.

The second case is also undefined, for the same reason the first one is
(repeated modifications of an object without an intervening sequence
point).
Christian

[CBFalconer]

ra*******@gmail .com wrote:

1) First how following program get executed i mean how output is
printed and also why following program gives different output in
Turbo C++ compiler and Visual c++ 6 compiler?
void main()
{
int val=5;
printf("%d %d %d %d",val,--val,++val,val--);
}
under turbo compiler its giving
4 4 5 5
and under visual c++ its
5 5 6 5

2) How to evaluate following statement
int val =5;
val =- --val - val-- - --val;

Get your C book or the standard and read up on sequence points
and/or parameter evaluation order. This is elementary. And if you
really mean C++ this is not the place.

--
"If you want to post a followup via groups.google.c om, don't use
the broken "Reply" link at the bottom of the article. Click on
"show options" at the top of the article, then click on the
"Reply" at the bottom of the article headers." - Keith Thompson

[akarl]

John Bode wrote:

ra*******@gmail .com wrote:

hello,
1) First how following program get executed i mean how output is
printed and also why following program gives different output in Turbo
C++ compiler and Visual c++ 6 compiler?

It's because you are invoking undefined behavior in both cases. Read
the following: http://www.eskimo.com/~scs/C-faq/s3.html

The expression i++ evaluates to the current value of i; the side affect
(i incremented by 1) is applied sometime before the next sequence
point, but it need not be immediately after the expression is
evaluated.

Similarly for ++i; it evaluates to the current value of i + 1, but i is
not necessarily incremented immediately.

There is no requirement on when the side affects are applied other than
it must happen before the next sequence point.

For example, given the statement

i = j++ + ++k;

the following sequence is possible:

t1 <- j
t2 <- k + 1
i <- t1 + t2
j <- j + 1
k <- k + 1

Same thing applies to your printf() statement; there's no sequence
point between the autoincrement/decrement expressions, so the behavior
is undefined, meaning the compiler can do anything it wants.

void main()

You mean int main(void). Strictly speaking, "void main()" is an error.

{
int val=5;
printf("%d %d %d %d",val,--val,++val,val--);
}
under turbo compiler its giving
4 4 5 5
and under visual c++ its
5 5 6 5

2) How to evaluate following statement
int val =5;
val =- --val - val-- - --val;

How come statements like these compile when the result is undefined? Is
it too hard for the compiler to figure out that it results in undefined
behavior?

[Richard Bos]

akarl <fu********@com hem.se> wrote:

John Bode wrote:

ra*******@gmail .com wrote:

2) How to evaluate following statement
int val =5;
val =- --val - val-- - --val;

How come statements like these compile when the result is undefined? Is
it too hard for the compiler to figure out that it results in undefined
behavior?

In a simple case like this, perhaps not; but in the general case, yes.

Richard

[John Bode]

akarl wrote:

John Bode wrote:

ra*******@gmail .com wrote:

hello,
1) First how following program get executed i mean how output is
printed and also why following program gives different output in Turbo
C++ compiler and Visual c++ 6 compiler?

It's because you are invoking undefined behavior in both cases. Read
the following: http://www.eskimo.com/~scs/C-faq/s3.html

[snip]

How come statements like these compile when the result is undefined?
Well, "undefined behavior" basically means the compiler is free to
handle the problem in any way it wants to. As soon as the Standard
mandates that the compiler issue a diagnostic or abort, the behavior is
no longer undefined.
Is
it too hard for the compiler to figure out that it results in undefined
behavior?

I'm not a compiler writer by trade, so I can't say for sure. In the
obvious case (i = i++), it shouldn't be too hard. But imagine a case
like this:

foo.c
-----------------------
int foo(int *p, int *q)
{
return (*p)++ + ++(*q);
}

main.c
-----------------------
extern int foo(int *p, int *q);

int main(void)
{
int i, *j, *k;
int x;

k = &i;
i = 5;
j = k;

x = foo(j, k);
return 0;
}

Not only is i aliased by pointers, the foo() function is compiled
separately from the main() function. There's simply no way to detect
that i is being modified more than once between sequence points at
compile time.

Situations like that would be impossible to guard against. That's why
the Standard leaves the behavior "undefined" ; there's simply no good
way to catch all instances of this problem at compile time, so no
requirement is placed on the compiler implementor to do so.