Value, value, who's got the value?

[Tim Olson]

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;

The OS doesn't cause the difference -- the compiler does.  Since func1
and func2 have no return statements, the value returned is whatever last
happend to be placed in the return-value location. On your UNIX-PC, this
is probably the same register as the first "temp" register that gets
assigned during expression evaluation.  On the BSD machine, it isn't.

| why does
| System V get it 'right' (even tho the code is wrong);

The compilers on that machine under that OS just happened to do what you
thought should happen.  The code is wrong, because there are no explicit
return statements.

| why do
| none of these flag func2 as having a syntax error?

Because it isn't a syntax error -- it is a legal (although useless)
expression.  Remember that in C, assignments aren't statements -- they
are expressions.

Here is what our local C compiler had to say about your program:

w "t.c",L1/C9:  func1: Function has no return statement.
w "t.c",L12/C9: c: Expression has no side-effects.
w "t.c",L8/C9:  func2: Function has no return statement.
w "t.c",L18/C9: printf: Function called but not defined.


	-- Tim Olson
	Advanced Micro Devices
	(tim at amd.com)