if(a,b,c)

[hyderabadblues]

Why the following statement evaluates as false infact it doesn't follow
the syntax

int a=10,b=8,c=0;

if(a,b,c)
printf("=");
else
printf("!=");

prints !=

[Kenneth Brody]

hyderabadblues wrote:

Why the following statement evaluates as false infact it doesn't follow
the syntax

int a=10,b=8,c=0;

if(a,b,c)
printf("=");
else
printf("!=");

prints !=

Because c==0, therefore (a,b,c)==0, therefore "if (a,b,c)" is false.

And what syntax does it not follow?

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[Eric Sosman]

hyderabadblues wrote On 10/11/05 13:54,:

Why the following statement evaluates as false infact it doesn't follow
the syntax

int a=10,b=8,c=0;

if(a,b,c)
printf("=");
else
printf("!=");

prints !=

Search your C textbook for "the comma operator."

--
Er*********@sun.com

[Martin Ambuhl]

hyderabadblues wrote:

Why the following statement evaluates as false infact it doesn't follow
the syntax
What do you mean "it doesn't follow the syntax"?
int a=10,b=8,c=0;

if(a,b,c)
The commas are comma operators (what did you think they were?), so the
above is just
if (c)
with the side effect of evaluating a and b along the way.
Since (c == 0), the condition is never true, so ...
printf("=");
else
printf("!=");

prints !=
of course.

And next time please post compilable code.

[Charles M. Reinke]

"hyderabadblues" <si*********@gmail.com> wrote in message
news:11**********************@f14g2000cwb.googlegr oups.com...

Why the following statement evaluates as false infact it doesn't follow
the syntax

int a=10,b=8,c=0;

if(a,b,c)
printf("=");
else
printf("!=");

prints !=

IIRC, the comma operator results in the evaluation of all arguments, then
returns the value of the LAST (i.e. rightmost) argument. In this case, it
evaluates the expressions "a", "b", and "c" and then returns the value of
the last expression ("c"), which is 0. Thus, the correct behavior is for
the "if" statement to be false, and the "else" block to be executed, as you
observed. What were you expecting it to do?

-Charles

[Default User]

hyderabadblues wrote:

Why the following statement evaluates as false infact it doesn't
follow the syntax
Yes it does.
int a=10,b=8,c=0;

if(a,b,c)

Look in your book to see how the comma operator works.

Brian

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[Mike Wahler]

"hyderabadblues" <si*********@gmail.com> wrote in message
news:11**********************@f14g2000cwb.googlegr oups.com...

Why the following statement evaluates as false
Because 'c' evaluates to zero, and zero always
evaluates to false (non-zero always evaluates to
true).
infact it doesn't follow
the syntax
Sure it does. It's valid C syntax anyway.
You just don't appear to understand what
that syntax does.

Get a C book (or more).

-Mike

int a=10,b=8,c=0;

if(a,b,c)
printf("=");
else
printf("!=");

prints !=

[g.kanaka.raju]

hyderabadblues wrote:

Why the following statement evaluates as false infact it doesn't follow
the syntax

int a=10,b=8,c=0;

if(a,b,c)
printf("=");
else
printf("!=");

prints !=

The syntax is fine and the behaviour of the program as per the
semantics of "," operator. It's associativity is left-to-right and if a
pair of expressions seperated by a comma is evaluated left-to-right.
So, there's no ambiguity even with the behaviour as well. So, the
program always prints !=.

Regards,
Raju