Value, value, who's got the value?

[Jeremy Roussak]

In article <1044 at itivax.iti.org> scs at vax3.iti.org (Steve Simmons) writes:
>
>Consider the following program:
>
>int	func1()
>{
>	int b ;
>
>	b = 2 ;
>}
>
>int	func2()
>{
>	int c = 3 ;
>
>	c ;
>}
>
>main()
>{
>	int a = 1 ;
>	printf( "Value of a is %d\n", a ) ;
>	a = func1() ;
>	printf( "Value of a is %d\n", a ) ;
>	a = func2() ;
>	printf( "Value of a is %d\n", a ) ;
>}
>
>Compile and run this on a UNIX-PC (system V) under standard cc or
>with gcc, and the result is:
>	Value of a is 1
>	Value of a is 2
>	Value of a is 3
>
>On BSD43. with standard cc or gcc, the result is
>	Value of a is 1
>	Value of a is 0
>	Value of a is 0
>
>Several questions: why does the OS make a difference; why does
>System V get it 'right' (even tho the code is wrong); why do
>none of these flag func2 as having a syntax error?

To answer your questions in reverse order, no syntax error is flagged
because there is no syntax error.  A statement in C is simply an expression.
The statement a+3; is perfectly legal: the compiler will add 3 to a and
blissfully throw the result away.  Think of the "statement" i++;

It's not so much the  OS that makes the difference as the way the compiler
generates code.  If it leaves values used in assignments lying around
in a place where they might be interpreted later as results from a function,
and your function carelessly doesn't define a return value, you'll
get a spurious return value.  In your case, this is the value you want
but haven't written the code to get.  To choose a silly example, some
68000 C compilers use d0 to return function results.  If an assignment
   a=3
generated code like
   move #3,d0
   move d0,a   (not that it would, one hopes!)
then the value 3 would be lying around in d0 at return time.

Simple, eh?

Jeremy Roussak       Just a part-time hacker