Question

Related question: Any good reason why assignment operator isn't a sequence point?

From the comp.lang.c FAQ I would infer that the program below is undefined. Strangely, it only mentions the call to f as a sequence point, between the computation of the arguments and the transfer of control to f. The transfer of control from f back to the calling expression is not listed as a sequence point.

int f(void) { i++; return 42; }
i = f();

Is it really undefined?

As an end-note that I add to many of my questions, I am interested in this in the context of static analysis. I am not writing this myself, I just want to know if I should warn about it in programs written by others.

Answer 1

The transfer of control from f back to
the calling expression is not listed
as a sequence point.

Yes it is.

at the end of the evaluation of a full expression

 

The complete expression that forms an
expression statement, or one of the
controlling expressions of an if,
switch, while, for, or do/while
statement, or the expression in an
initializer or a return statement.

You have a return statement, therefore, you have a sequence point.

It doesn’t even appear that

int f(void) { return i++; } // sequence point here, so I guess we're good
i = f();

is undefined. (Which to me is kind of weird.)