Alex Rivera | Logout

Why are these constructs using pre and post-increment undefined behavior?

Asked 2009-06-04T09:17:52.267
927
#include <stdio.h>

int main(void)
{
   int i = 0;
   i = i++ + ++i;
   printf("%d\n", i); // 3

   i = 1;
   i = (i++);
   printf("%d\n", i); // 2 Should be 1, no ?

   volatile int u = 0;
   u = u++ + ++u;
   printf("%d\n", u); // 1

   u = 1;
   u = (u++);
   printf("%d\n", u); // 2 Should also be one, no ?

   register int v = 0;
   v = v++ + ++v;
   printf("%d\n", v); // 3 (Should be the same as u ?)

   int w = 0;
   printf("%d %d\n", ++w, w); // shouldn't this print 1 1

   int x[2] = { 5, 8 }, y = 0;
   x[y] = y ++;
   printf("%d %d\n", x[0], x[1]); // shouldn't this print 0 8? or 5 0?
}
c increment undefined-behavior operator-precedence sequence-points
Edit
Report

1 Answer

23

While it is unlikely that any compilers and processors would actually do so, it would be legal, under the C standard, for the compiler to implement "i++" with the sequence:

In a single operation, read `i` and lock it to prevent access until further notice
Compute (1+read_value)
In a single operation, unlock `i` and store the computed value

While I don't think any processors support the hardware to allow such a thing to be done efficiently, one can easily imagine situations where such behavior would make multi-threaded code easier (e.g. it would guarantee that if two threads try to perform the above sequence simultaneously, i would get incremented by two) and it's not totally inconceivable that some future processor might provide a feature something like that.

If the compiler were to write i++ as indicated above (legal under the standard) and were to intersperse the above instructions throughout the evaluation of the overall expression (also legal), and if it didn't happen to notice that one of the other instructions happened to access i, it would be possible (and legal) for the compiler to generate a sequence of instructions that would deadlock. To be sure, a compiler would almost certainly detect the problem in the case where the same variable i is used in both places, but if a routine accepts references to two pointers p and q, and uses (*p) and (*q) in the above expression (rather than using i twice) the compiler would not be required to recognize or avoid the deadlock that would occur if the same object's address were passed for both p and q.

answered 2012-12-05T18:30:27.523

Your Answer