Thanks to Sam for the idea.
int k = 0; k = k++;
After both lines execute, what is k equal to?
Apparently, the answer depends on a number of things. In Java, k is 0. In C, it may be 0 or 1 depending on your compiler!
Bonus points if you can explain why.
The system/application admin says:
Because some parsers see ++ as a sequential step to perform after the items before it.
“the variable k is equal to the variable k. now increment k by one.”
In some languages this can be solved by writing the code like this:
k = ++k;
Which translates to “the variable k is equal to the incremented value of k.”
Or I could be completely wrong… :-)
I would guess that the result in C was dependent on how closely the compiler follows operator precedence/binding.
The postfix increment operator “++” is at the top of the precedence chart where direct assignment “=” is almost at the bottom.
Wouldn’t CPP turn it into:
k = (k++)
and not:
k = k;
k++;
??
Then I found this:
“Postfix increment/decrement have high precedence, but the actual increment or decrement of the operand is delayed (to be accomplished sometime before the statement completes execution). So in the statement y = x * z++; the current value of z is used to evaluate the expression (i.e., z++ evaluates to z) and z only incremented after all else is done.”
I couldn’t locate a copy of the ANSI C standard in the 5 min. I spent looking…
YMMV. Darn!