🚀 go-pugleaf

In article <11900001@maggot> barada@maggot.applicon.UUCP writes:

>> >	a = ((b=1),b) + ((b=2),b) + ((b=3),b)
>
>I rearraged the above to
>
>	a = (b,(b=1)) + (b,(b=2)) + (b,(b=3));

   this isn't a rearragement (sic), but a different statement.
>
>And my BSD4.2 VAX produced:
>
>        movl    $1,_b
>        movl    _b,r0
>        movl    $2,_b
>        movl    _b,r1
>        addl2   r1,r0
>        movl    $3,_b
>        movl    _b,r1
>        addl2   r1,r0
>        movl    r0,_a
>									 
>As you can see, the comma operator is evaluated right to left. I think that

No, it is not.

>this is a serious bug.

Ah yes, but the bug is in your code, not the compiler.
>
>BTW, this code produces the proper answer of 6.
>
There is no single proper answer.

The comma operator is defined to evaluate the left operand, discard it,
then evaluate the right operand, and return the resulting value, ie
that of the RIGHT operand.  So,
   (b,b=1)
evaluates b and discards it, evaluates b=1, assigning one to b, and returns
the result which is the new value of b.  Your statement is therefore
equivalent to
	a = (b=1) + (b=2) + (b=3);
which is equiivalent to
	a = 1 + 2 + 3;
followed or preceded by the assignements to b in ANY order.
The compiler has elided the evaluations of b, as the result is to be
discarded and there are no side-effects.

In summary, the compiler *is* allowed to reorder operands of operators
like +, but you are *not* allowed to reorder the operands of ","
and expect the same result.
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