cant understand this output?

[tijoriwalaritesh]

Below is a simple program.

int i = 0;
printf("%d %d %d %d", i++, i==1, i, ++i)

Its giving this output:
On Sun Solaris 8:
0 1 2 2
On Linux:
1 1 1 1
On Windows:
0 1 1 2

Can any one explain me how the compiler behaves in any particular case. I know the code is compiler dependent. Lets assume on linux.

also check your explanation with this one:

int i = 0;
printf("%d %d %d %d", (i++) + (++i), i==1, i, ++i)

 

[cpjust]

The Windows output is correct. I don't have a clue what's going on with UNIX though?

 

[ArkM]

Indeed you have an incorrect statement:

printf("%d %d %d %d", i++, i==1, i, ++i)

See

Certain other aspects and operations of the abstract machine are described in this International Standard as unspecified (for example, order of evaluation of arguments to a function). ...

Incr/decr ops have side effect so arg list values are undefined in this case. It's an excellent example of a classic trap with evaluation order. You may have different results not only on Unix/Windows but also on different implementation on the same system.
Be careful with side effect operations.

 

[ArkM]

Moreover: result of incr/decr op is in effect after full arg list evaluation, so you may have i == 0, 1 or 2 in the moment of i == 1 evaluation.

 

[Salem]

> printf("%d %d %d %d", i++, i==1, i, ++i)
They're all right, and they're all wrong.

http://c-faq.com/expr/evalorder1.html

Multiple side effects on the SAME variable between sequence points are ALWAYS UNDEFINED within the C language.

Basically, the code is broken.


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